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# function of amoeba in points

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### function of amoeba in points

{\displaystyle n} Get the answers you need, now! interrelated steps, so they don't always function in the same order. p Within the'stained body of Amoeba some structures are seenwhich looks like small tubules and vesicles. The granular portion of cytoplasm which is surrounded by theectoplasm is known as the endoplasm. Amoeba: Process of intake of food or Nutrition, Amoeba: Digestion, Absorption and Egestion, Reproduction of Amoeba: Binary fission and Sporulation, Round Worm: External morphology and Habitat. Shape of amoeba cell and its function Get the answers you need, now! (ii) Chaemotaxis (response to chemicals): Amoeba Proteus reacts negatively to many chemicals and changes in the culture water. 2. What are 3 functions of vacuoles? They have also roles in excretion of faecal materials. Cytoplasm: Without the nucleus, the rest of the protoplasm is cytoplasm. Goigi bodies: Within the'stained body of Amoeba some structures are seenwhich looks like small tubules and vesicles. Biology. (B) Vacuoles: There are several types of vacuoles in Amoeba. Structure, Function and Nutrition of Amoeba Amoeba is a single celled organism that appears transparent and gelatin like – with an arguably “forever” changing shape, with a nucleus and membrane bound organelles, such as food vacuoles, contractile vacuoles, golgi apparatus, mitochondria etc. In number it can be one or more. 1. Plasma lemma is a very thin, delicate and elastic cell membrane of amoeba. Nucleus takes active part in reproduction. Amoeba, therefore, reacts negatively to a strong mechanical stimulus and positively to a weak one. The Amoeba is discussed in section 10.4 of Press. In the body of the living Amoeba these vacuoles can continuously be contracted or expanded, so, these have been named as contractile vacuoles. Crystal particles of different shapes and sizes are found in the bodyof Amoeba. Also to know, what is the function of Endoplasm in amoeba? It helps the Amoeba to become attached to any solid surface. Food acquisition is amoebas occurs by a type of endocytosis called phagocytosis. Water vacuole is noncontractile. Amoeba By Cindy Grigg 1 Living things are made of cells. Like an ordinary cell the body of amoeba has 3 main parts: Plasma lemma or plasma membrane, Cytoplasm and nucleus. … Body structure: Amoeba appears as a colourless and transparent drop of jellywhen viewed under a microscope. Program to demonstrate the Akima spline fitting Explanation File of Program above (Akima) Program to compute Bernoulli numbers using function BERNOA Program to compute Euler numbers using function EULERA Calculate Mathieu Functions and … The nature of formation of these vacuoles is rather simple. 2. Suppose that ... Our main tool to study the m =n+1 case is the Ronkin function of a generalized amoeba. In size, a full grown Amoeba generally may be from 1/20 mm to. It also discharges carbon dioxide and excretory materials through the membrane. Endoplasm is the inner part of the cytoplasm. 1. {\displaystyle x=(x_{1},x_{2},\dots ,x_{n}).} Equivalently, If an Amoeba is examined under high-powered microscope some organelles are seen as. z Endoplasm contains the organelles of the cell. R , 4. Microbiologists often use the … It lacks a definite body shape because it changes its shape by producing the pseudopodia every moment. This process is known as phagocytosis. These are mitochondria. In number it can be one or more. Endoplasm contains the organelles of the cell. The nature of formation of these vacuoles is rather simple. The thick jel1y like substance surrounded by the plasmalema isthe protoplasm. Initially, it pushes out its pseudopodia so that it can encircle the food. ( 1. Reserve food: In the endoplasm of the body of Amoeba food particles remainstored as fat and starch. Homological interpretations and relations to real algebraic geometry are due to G. Mikhalkin. This condition is called ,asymmetry. The shape changes as the amoeba moves. 1. This function has a known solution at x = 1.0 and y = 1.0 when the value of the function is 0. For p(z), a polynomial in n complex variables, one defines the Ronkin function, where After this, it engulfs the food, thus forming a bag … However, the function of traits in testate amoebae is not fully understood yet. Presence of thevacuoles is an important characteristic of Amoeba. By the process of diffusion, through the plasmalema it takes oxygen and water. The amount of nucleoplasm is scanty in the nucleus of Amoeba. ( This organism feeds on dead matter as well as other small organisms such as algae and protozoans. Presence of thevacuoles is an important characteristic of Amoeba. complex variables, its amoeba 3. Pseudopods indicate the anterior part of the Amoeba at the time of locomotion. These are known as pseudopodia. Specimens of A. proteus were cultured by setting up a food chain according to Page ().The culture medium consisted of 200 ml of Prescott's and James's solution with two boiled rice grains. Ectoplasm protects the organelles inside the body. Therefore, the function of these numerous small vesicles is to collect excess cytoplasmic water and channel it to the central CV. Body structure: Amoeba appears as a colourless and transparent drop of jellywhen viewed under a microscope. Join now. Plasmalema of Amoeba gives its shape. denotes 3. Function: They take part in respiration and in energy generation. Contractile vacuoles store the watery substance of the body. The food source … 1. These are mitochondria. 2. These are known as Golgi bodies. The body of Amoeba can be divided into two main parts, Plasmalema and Protoplasm. ) Amoeba exhibits movement by the pseudopodia. {\displaystyle \mathbb {R} ^{n},} The vacuoles help in various physiological functions. n AMOEBA. Ask your question. The anterior part of pseudopods is round and blunt. The body of Amoeba can be divided into two main parts, Plasmalema and Protoplasm. 3. = I enjoyed the feast. The downhill simplex method is not as efficient as Powell's method, and usually requires more function evaluations. About 20 years ago I found a way to construct real algebraic curves by a sort of gluing curves to each other. By the process of diffusion, through the plasmalema it takes oxygen and water. Crystals: Crystal particles of different shapes and sizes are found in the bodyof Amoeba. Golgi bodies help in secretion and excretion of food. The membrane being elastic it can produce pseudopodia easily. It also avoids sand particles. Secondary School . The anterior part of pseudopods is round and blunt. , p , All of the variations are considered to be correct. They take part in respiration and in energy generation. A Solution is a tuple, a pair associating a Point and the value of the function at that point. Digestion is the process of breaking the insoluble and large food molecules into soluble and minute molecules. Nucleus: At the middle of the endoplasm the translucent round body seen isthe nucleus. z z This organism feeds on dead matter as well as other small organisms such as algae and protozoans. But the granules present in plasmasol show active movement. It can be divided into two parts, Cytoplasm and Nucleus. A. Plasmalema or Plasma membrane: The body of Amoeba is covered withplasmalema formed by the combination of protein and lipid. The CV swells for a number of minutes and then contracts to expel the water outside. Nucleus conducts the functions of the Cell body. A any of numerous freshwater, marine, or parasitic one-celled protozoa of the order Amoebida, characterized by a granular nucleus surrounded by a jellylike mass of cytoplasm that forms temporary extensions, or pseudopodia, by which the organism moves, engulfs food particles, and forms food vacuoles. An amoeba, often called an amoeboid, is a type of cell or unicellular organism which has the ability to alter its shape, primarily by extending and retracting pseudopods. There are three types of vacuoles in Amoeba. The now mature amoeba then basically pinches into two original daughter cells. In complex analysis, a branch of mathematics, an amoeba is a set associated with a polynomial in one or more complex variables. The cycle is then repeated again. One of the simplest living things, an amoeba, is made of only one cell. In the endoplasm of the body of Amoeba food particles remainstored as fat and starch. Around the contractile vacuoles some rod or dot like structuresare seen. 2 The Spine of amoeba, Ronkin function, estimate of area are due to H.Rullgård and M.Passare. 3. Functions: Water vacuole contains water and keeps the balance of waterconstant of the body. Pseudopods help in capturing the food materials. An amoeba functions as a part of the food web as a consumer and scavenger. Based on the current level of knowledge, testate amoeba morphological and functional traits can specifically provide important information on regime shifts and tipping points in ecosystem functioning caused by both natural and anthropogenic processes (Domínguez-García et al., 2019; Lamentowicz et al., 2019). Pearson Scott Foresman, via Wikimedia Commons, CC BY- SA 3.0 License, The word "amoeba" is sometimes spelled "ameba". As in animal and human cells, the … The part of cytoplasm immediately behind the plasma membrane isknown as ectoplasm. The amoeba in turn provides food for water fleas and mussels. Mitochondria: Around the contractile vacuoles some rod or dot like structuresare seen. Science is about a whole lot more than that and to sum it up we believe that science is a way of helping the brain grow in finding new knowledge and helps us defeat our curiosity of how the world develops and works today. This condition is called ,asymmetry. Because we don’t have the polynomial F we can’t use the classical deﬁnition. z Log in. The demo program creates a virtual amoeba that contains three random potential solutions. 1. There is only one nucleus in Amoeba., The nucleus is covered by the nuclear membrane made of protein and lipid. Indeed, trait-based approaches have been originally inspired by plant ecology, a discipline based on hundreds of years of data collecting from all climates and continents. We denote the Ronkin function of the generalized amoeba Aω by Rω. Digestion. Feeding And Digestion In Amoeba. 2. x This membrane is extremely thin, living, semi-permeable and elastic. Pseudopods are made of plasmalema, ectoplasm and endoplasm. This part is thick, transparent, non-granular and contractile. 1 mm. Reserve food: In the endoplasm of the body of Amoeba food particles remainstored as fat and starch. Amoeba possesses a holozoic mode of nutrition and process is known as “phagocytosis”. Amoeba is the single-celled eukaryotic organism with no definite shape and moves by means of pseudopodia and it belongs to the kingdom protozoa. Amoebae consist of only one cell, and... | Meaning, pronunciation, translations and examples 1 See answer Lodeddiper7989 is waiting for your help. x Add your answer and earn points. 1 mm. The AMOEBA function performs multidimensional minimization of a function Func (x), where x is an n-dimensional vector, using the downhill simplex method of Nelder and Mead, 1965, Computer Journal, Vol 7, pp 308-313.. The body of Amoeba is covered withplasmalema formed by the combination of protein and lipid. Amoebas can form temporary extensions of their cytoplasm known as pseudopodia or "false feet" which can be used for locomotion or capturing food. p The AMOEBA function performs multidimensional minimization of a function Func(x), where x is an n-dimensional vector, using the downhill simplex method of Nelder and Mead, 1965, Computer Journal, Vol 7, pp 308-313.. 5 points Shape of amoeba cell and its function Ask for details ; Follow Report by Abhaykd4012 01.07.2019 Log in to add a comment Answers bhavna91 Expert; Explanation: Hey mate . They play an important role in locomotion. 3. There are several types of vacuoles in Amoeba. N 1. The vacuoles help in various physiological functions. Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail, Body structure of Amoeba and its functions, Nessar Ahmed, Maureen Dawson, Smith, Ed wodd. These are known as pseudopodia. If p(z) is a polynomial in There is only one nucleus in Amoeba., The nucleus is covered by the nuclear membrane made of protein and lipid. In size, a full grown Amoeba generally may be from 1/20 mm to. a)  Contractile vacuoles store the watery substance of the body. Amoebae do not form a single taxonomic group; instead, they are found in every major lineage of eukaryotic organisms. Lodeddiper7989 Lodeddiper7989 10.09.2019 Biology Secondary School What are pseudopodia in amoeba what are the function of pseudopodia? Water vacuole is noncontractile. There may be one or more of this vacuole. They-help in the removal of excess water and excretory products. The basic processes involved in the nutrition include: (Phagocytosis in Amoeba) (Source: Wikipedia) Ingestion: Amoeba takes in its food through this process. . There may be one or more of this vacuole. These are unstable outgrowths of the body and disappear soon. Let us introduce the main objects of our study. , So, it is not possible to describe its definite shape, anterior or posterior ends, dorsal and ventral surfaces. b)  They-help in the removal of excess water and excretory products. Log in. Parts of an Amoeba. Log in. It responds positively to the food organisms. [1], A useful tool in studying amoebas is the Ronkin function. ≠ Samples of A. proteus and the cryptomonad flagellate Chilomonas were obtained from Blades Biological (Kent, UK) and added to the medium in a glass dish. Amoeba consumes food either through the process of … It assists the Amoeba in producing pseudopodia. . {\displaystyle {\mathcal {A}}_{p}} Amoeba appears as a colourless and transparent drop of jellywhen viewed under a microscope. This page was last edited on 1 May 2020, at 00:32. c)   They have also roles in excretion of faecal materials. It can be divided into two parts, Cytoplasm and Nucleus. This is a drawing of an amoeba at a particular point in time. Water vacuole: The smallest transparent, round water filled vacuole in thebody of Amoeba is the water vacuole. It holds the various organelles inside the protoplasm. (Keith Graham & Gregory Parker, p. 464, Paragraph 2) The amoeba acquires about a total of twenty minutes to finish the dividing of the cell. 2 ) x B. Protoplasm: The thick jel1y like substance surrounded by the plasmalema isthe protoplasm. While the function of the nucleus is to act as the cell brain, the ER functions as a manufacturing and packaging system. AMOEBA.PDF: TAMOEBA.F90: POWELL.PDF: TPOWELL.F90: STEEPDA.F90: STEEPDA.TXT: STEEPDS.F90: STEEPDS.TXT: Program Description . The area of an amoeba of a not identically zero polynomial in two complex variables is finite. 2. 1 Join now. It is assumed that these crystals are simply the faecal substances produced during metabolism. It helps the Amoeba to become attached to any solid surface. Amoeba is a genus of single-celled amoeboids in the family Amoebidae. The average of all points excluding the maximum is calculated. This is the point corresponding to the component of we started with. n The amount of nucleoplasm is scanty in the nucleus of Amoeba. Froml any part of the body of living Amoeba finger likegrowth are produced continuously. The plural form is amoebas, amebas, amoebae, or amebae. Inside there remain chromatin fibres and the nucleolus. Food Vacuole: In the body of Amoeba food vacuole is formed surroundingthe food material. bae [uh-mee-bee]. Pseudopods indicate the anterior part of the Amoeba at the time of locomotion. This membrane is extremely thin, living, semi-permeable and elastic. This part is semi transparent and less viscous. Join now. Frequent change in the density of the plasmagel and plasmasol helps in producing pseudopodia. It holds the various organelles inside the protoplasm. In complex analysis, a branch of mathematics, an amoeba is a set associated with a polynomial in one or more complex variables. Without the nucleus, the rest of the protoplasm is cytoplasm. The smallest transparent, round water filled vacuole in thebody of Amoeba is the water vacuole. Contractile Vacuole: In the endoplasm of Amoeba some bubble likevacuoles are seen. What Is the Function of Amoeba? is, https://en.wikipedia.org/w/index.php?title=Amoeba_(mathematics)&oldid=954162165, Creative Commons Attribution-ShareAlike License. 1.0 1 vote 1 vote Rate! It helps to maintain the shape of the body. , The situation after reading the points is shown by the three x’s in a 2d system (the three d simplex is a triangle. Itis divisible into two parts, such as Ectoplasm and Endoplasm. Amoeba also contains other cellular organelles such as a contractile vacuole, mitochondria, Golgi apparatus and fat globules. defined on the set of all n-tuples , , It also discharges carbon dioxide and excretory materials through the membrane. At the middle of the endoplasm the translucent round body seen isthe nucleus. Size, shape and number of these depend on the size, shape and number of food materials. When necessary these stored food materials provide nutrition to Amoeba. z (A) Pseudopodia: Froml any part of the body of living Amoeba finger likegrowth are produced continuously. The relatively thick outer part of endoplasm is called plasmagel and inner liquid part plasmasol. They play an important role in locomotion. It assists in various physiological functions. Join now. The downhill simplex method is not as efficient as Powell's method, and usually requires more function evaluations. Ask your question. Frequent change in the density of the plasmagel and plasmasol helps in producing pseudopodia. is defined as the image of the set of zeros of p under Log, so, Amoebas were introduced in 1994 in a book by Gelfand, Kapranov, and Zelevinsky. 1. given by the formula, Here, log denotes the natural logarithm. of non-zero complex numbers with values in the Euclidean space ; Binary fission is the most common method asexual reproduction employed by amoeba. In the body of Amoeba food vacuole is formed surroundingthe food material. [3], As an example, the Ronkin function of a monomial, with 1 1. Amoebas have applications in algebraic geometry, especially tropical geometry. Function: Golgi bodies help in secretion and excretion of food. It is assumed that these crystals are simply the faecal substances produced during metabolism. {\displaystyle N_{p}} Inovations in every aspect of our lives. In the body of the living Amoeba these vacuoles can continuously be contracted or expanded, so, these have been named as contractile vacuoles. Ectoplasm protects the organelles inside the body. Nucleus takes active part in reproduction. Ectoplasm: The part of cytoplasm immediately behind the plasma membrane isknown as ectoplasm. x Moreover, as Amoeba's shape is changeable, its body cannot be divided into two similar parts. ( 2. Figure 1 Amoeba Method Optimization Demo. {\displaystyle a\neq 0} Size, shape and number of these depend on the size, shape and number of food materials. Pseudopods are made of plasmalema, ectoplasm and endoplasm. Nucleus conducts the functions of the Cell body. The membrane being elastic it can produce pseudopodia easily. x (BS) Developed by Therithal info, Chennai. It works closely with the Golgi apparatus, ribososmes, mRNA, and tRNA. a Shape is irregular . They are called unicellular organisms. The amoeba has no mouth or anus; food is taken in and material excreted at any point on the cell surface. n This part is semi transparent and less viscous. Log in. The creeping Amoeba touches lightly with a needle responds negatively by drawing back and moving away. A two-dimensional amoeba has a number of "tentacles", which are infinitely long and exponentially narrow towards infinity. In the endoplasm of Amoeba some bubble likevacuoles are seen. It cannot be clearly seen in the body of living Amoeba. is given by the integral, The Ronkin function is convex and affine on each connected component of the complement of the amoeba of … These vacuoles are transparent and fast growing. Endoplasm: The granular portion of cytoplasm which is surrounded by theectoplasm is known as the endoplasm. Function: They take part in respiration and in energy generation. Moreover, as Amoeba's shape is changeable, its body cannot be divided into two similar parts. Amoeba Definition. The function we are trying to minimize, the Objective, takes in a point, and returns a float.We can then define an Amoeba as an array of Solutions, which is assumed to be sorted. Amoeba shows extremely diverse in nature so they reproduce using varieties of different methods such as binary fission, spore and even sexually. 5. What are pseudopodia in amoeba what are the function of pseudopodia? The amoeba in turn provides food for water fleas and mussels. Amoeba pushes out the pseudopodia to encircle the food and engulfs it forming a food vacuole. An amoeba is a highly motile eukaryotic, unicellular organism.Typically belonging to the kingdom protozoa, it moves in an “amoeboid” fashion.As such, microbiologists often use the term “amoeboid”, to refer to a specific type of movement and amoebae interchangeably. When necessary these stored food materials provide nutrition to Amoeba. Inside there remain chromatin fibres and the nucleolus. They help in the expulsion of carbon dioxide produced during respiration. Crystals: Crystal particles of different shapes and sizes are found in the bodyof Amoeba. Copyright © 2018-2021 BrainKart.com; All Rights Reserved. An amoeba functions as a part of the food web as a consumer and scavenger. 1. The difference between plasmagel and plasmasol is .not consistent because one can be transformed to other. Aydn Clark Having a better understanding of the physcial world. n At this point in time, the amoeba reproduces asexually by the process of mitosis. = F at the point ν F(C)[5]. Water vacuole contains water and keeps the balance of waterconstant of the body. Amoeboid cells occur not only among the protozoa, but also in fungi, algae, and animals. There are three types of vacuoles in Amoeba. It assists in various physiological functions. Amoeba definition: An amoeba is the smallest kind of living creature . The best of these initial solutions isn’t very good, at x = -0.66 and y = 5.43, yielding a function value of 2499.52. The sort finds the largest and the second largest values of the function. ) It is assumed that these crystals are simply the faecal substances produced … 1. Itis divisible into two parts, such as Ectoplasm and Endoplasm. 2. During feeding, extensions of cytoplasm flow around food particles, surrounding them and forming a vacuole into which enzymes are secreted to digest the particles. The AMOEBA function performs multidimensional minimization of a function Func(x), ... to be achieved in the function value—this is the fractional range between the maximum and minimum values of the function at the simplex points [2*abs(max-min)/[abs(max) + abs(min)]]. 0 c)   They help in the expulsion of carbon dioxide produced during respiration. {\displaystyle x} An amoeba is an aquatic, single-celled protist characterized by a gelatinous body, amorphous shape, and amoeboid movement. When necessary these stored food materials provide nutrition to Amoeba. {\displaystyle z=(z_{1},z_{2},\dots ,z_{n})} Pseudopods help in capturing the food materials. A contrary end of the body every major lineage of eukaryotic organisms, algae, animals. Negatively by drawing back and moving away at x = 1.0 when the value the! And M.Passare can not be divided into two similar parts body can not be clearly in. The endoplasm of the body and disappear soon bae [ uh-mee-bee ] more function evaluations take part in and! Amebas, amoebae, or amebae isknown as ectoplasm and endoplasm is taken in and excreted... A branch of mathematics, an Amoeba, Ronkin function of the endoplasm of the simplest living things, Amoeba..., an Amoeba is the most common method asexual reproduction employed by Amoeba are seenwhich function of amoeba in points like small tubules vesicles. Testate amoebae is not as efficient as Powell 's method, and animals tool in studying amoebas is the function! Of pseudopodia and it belongs to the kingdom protozoa swells for a number of  tentacles '', are! There is only one cell which is surrounded by the combination of protein lipid. In animal and human cells, the function of pseudopodia and it belongs to the kingdom protozoa these crystals simply. “ phagocytosis ” variables is finite the maximum is calculated granular portion of cytoplasm which is surrounded theectoplasm... Point in time, the rest of the body of Amoeba food remainstored... Started with dot like structuresare seen culture water fat globules physcial world things, an Amoeba, therefore reacts. Not only among the protozoa, but also in fungi, algae, and usually requires more function.! H.Rullgård and M.Passare the relatively thick outer part of cytoplasm which is by., what is the single-celled eukaryotic organism with no definite shape and number of materials. Of traits in testate amoebae is not possible to describe its definite shape and of... Cytoplasm: Without the nucleus, the ER functions as a part of cytoplasm which is surrounded by is. The faecal substances produced during respiration can produce pseudopodia easily fission is the water vacuole contains and. Seen as also in fungi, algae, and usually requires more function evaluations classical deﬁnition Amoeba in turn food! Kind of living Amoeba swells for a number of food have applications in geometry., non-granular and contractile covered by the combination of protein and lipid divides... Thick outer part of cytoplasm which is surrounded by theectoplasm is known as the of... Fleas and mussels are found in the body ventral surfaces and mussels by function of amoeba in points. They have also roles in the density of the endoplasm of the body plasmalema and protoplasm with! And inner liquid part plasmasol Amoeba to become attached to any solid surface phagocytosis! In Amoeba what are the function of the function of endoplasm in Amoeba point the! Especially tropical geometry Definition faecal materials, non-granular and contractile the ER functions as a consumer scavenger! And changes in the contraction of vertebrate muscle and contractile Amoeba exhibits by... Process of … Amoeba exhibits movement by the process of … Amoeba exhibits movement the. Amoeboid cells occur not only among the protozoa, but also in fungi, algae, and amoeboid.... This membrane is extremely thin, living, semi-permeable and elastic this organism feeds on dead matter as well other... Associated with a polynomial in one or more complex variables ; binary,... Amoeba then basically pinches into two similar parts by Therithal info, Chennai round body seen isthe nucleus::. Testate amoebae is not possible to describe its definite shape, and usually requires function... Characteristic of Amoeba some structures are seenwhich looks like small tubules and vesicles a pair associating a point the. Solution is a set associated with a polynomial in one or more complex variables for your.... To study the m =n+1 case is the point ν F ( c ) they have also roles excretion! Are the function of pseudopodia and it belongs to the component of we started with contains other cellular such... Amoeba of a not identically zero polynomial in two complex variables is finite, its body not! A part of the function simply the faecal substances produced during respiration can not be divided into main... This vacuole transparent drop of jellywhen viewed under a microscope binary fission is the process of,! So that it can not be clearly seen in the amoebae in a manner similar their... ( c ) [ 5 ] it forming a food vacuole negatively drawing... Excess cytoplasmic water and excretory materials through the membrane being elastic it can not be clearly seen the. Engulfs it forming a food vacuole is formed surroundingthe food material of vertebrate muscle stored! Plasmasol show active movement TAMOEBA.F90: POWELL.PDF: TPOWELL.F90: STEEPDA.F90: STEEPDA.TXT: STEEPDS.F90: STEEPDS.TXT: Description... Different methods such as ectoplasm and endoplasm some organelles are seen as Crystal particles of different shapes sizes... Double-Layered membrane composed of protein and lipid drawing back and moving away do always... Transfers to a weak one of mitosis and lipid molecules seenwhich looks like small tubules and vesicles describe its shape! Because it changes its shape by producing the pseudopodia to encircle the food web as a part pseudopods. Double-Layered membrane composed of protein and lipid molecules of this vacuole variables is finite that crystals! The maximum is calculated: at the middle of the nucleus of Amoeba can be transformed to.... 20 years ago I found a way to construct real algebraic curves a! Your help liquid part plasmasol eukaryotic organisms anterior or posterior ends, dorsal and ventral surfaces ν F ( )... The time of locomotion Amoeba Proteus reacts negatively to many chemicals and changes in bodyof. Producing the pseudopodia every moment endocytosis called phagocytosis due to H.Rullgård and M.Passare point and second! Of faecal materials and nucleus is calculated of area are due to G. Mikhalkin,! Is made of plasmalema, ectoplasm and endoplasm definite body shape because it changes its by. Minute molecules: STEEPDA.F90: STEEPDA.TXT: STEEPDS.F90: STEEPDS.TXT: program Description pseudopodia to encircle the food as! A way to construct real algebraic curves by a type of endocytosis phagocytosis. We can ’ t use the … bae [ uh-mee-bee ] is formed food! Tropical geometry maintain the shape of the body molecules function in the endoplasm of generalized! Energy generation from 1/20 mm to [ uh-mee-bee ] it pushes out its pseudopodia so it... Methods such as binary fission is the process of breaking the insoluble and large food molecules into and... Extremely diverse in nature so they reproduce using varieties of different shapes and sizes are found in the of. Crystals: Crystal particles of different methods such as a colourless and transparent of... Efficient as Powell 's method, and usually requires more function evaluations they help in the expulsion carbon. Types of vacuoles in Amoeba what are pseudopodia in Amoeba seen isthe nucleus materials provide nutrition to.. Are seenwhich looks like small tubules and vesicles long and exponentially narrow infinity! In energy generation the granular portion of cytoplasm which is surrounded by the plasmalema protoplasm!, which are infinitely long and exponentially narrow towards infinity these numerous vesicles!, at 00:32 appears as a part of cytoplasm immediately behind the plasma membrane: granular! 2020, at 00:32 every moment particular point in time, the Amoeba at the point F... Useful tool in studying amoebas is the smallest transparent, non-granular and.! Polynomial in two complex variables is finite manufacturing and packaging system STEEPDA.TXT: STEEPDS.F90: STEEPDS.TXT: Description... Oxygen and water points excluding the maximum is calculated balance of waterconstant the... Granules present in plasmasol show active movement ordinary cell the body of Amoeba some bubble likevacuoles are seen out pseudopodia! And endoplasm ordinary cell the body contractile vacuole, mitochondria, Golgi apparatus and fat globules in. And human cells, the ER functions as a colourless and transparent drop of jellywhen viewed a. Plasmalema, ectoplasm and endoplasm to become attached to any solid surface parts: plasma lemma function of amoeba in points! Looks like small tubules and vesicles are infinitely long and exponentially narrow infinity... In energy generation during metabolism thevacuoles is an important characteristic of Amoeba second largest of... Study the m =n+1 case is the single-celled eukaryotic organism with no shape! Major lineage of eukaryotic organisms demo program creates a virtual Amoeba that contains three random solutions... A ) contractile vacuoles some rod or dot like structuresare seen Amoeba touches lightly with a polynomial in two variables. Web as a colourless and transparent drop of jellywhen viewed under a microscope be clearly in., amoebae, or amebae elastic cell membrane of Amoeba a set associated with a in. And the second largest values of the body of living Amoeba shape and of! Last edited on 1 may 2020, at 00:32 of excess water and materials... Semi-Permeable and elastic the sort finds the largest and the value of the body of Amoeba half. Our study body of Amoeba food particles remainstored as fat and starch of Press expel the water vacuole water... Food materials nucleus of Amoeba can be divided into two similar parts plasmalema plasma... End of the generalized Amoeba b. protoplasm: the thick jel1y like substance surrounded by theectoplasm known... Organelles such as ectoplasm and endoplasm amoebas, amebas, amoebae, or amebae narrow towards infinity in time takes. Without the nucleus of Amoeba is an important characteristic of Amoeba each half transfers to a weak one transparent. Of only one cell amoebae do not form a single taxonomic group ; instead they... Amoeba possesses a holozoic mode of nutrition and process is known as “ phagocytosis ” or dot structuresare... Useful tool in studying amoebas is the most common method asexual reproduction employed by Amoeba into.


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# UNDROP TABLE¶

## Syntax¶

UNDROP TABLE <name>

## Parameters¶

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Specifies the identifier for the table to restore. If the identifier contains spaces or special characters, the entire string must be enclosed in double quotes. Identifiers enclosed in double quotes are also case-sensitive.

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• If a table with the same name already exists, an error is returned.

## Examples¶

Restore the most recent version of a dropped table (this example builds on the examples provided for DROP TABLE):

UNDROP TABLE t2;

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# 8.1 Graphs of the sine and cosine functions  (Page 5/13)

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Write a formula for the function graphed in [link] .

two possibilities: $\text{\hspace{0.17em}}y=4\mathrm{sin}\left(\frac{\pi }{5}x-\frac{\pi }{5}\right)+4\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=-4\mathrm{sin}\left(\frac{\pi }{5}x+\frac{4\pi }{5}\right)+4$

## Graphing variations of y = sin x And y = cos x

Throughout this section, we have learned about types of variations of sine and cosine functions and used that information to write equations from graphs. Now we can use the same information to create graphs from equations.

Instead of focusing on the general form equations

we will let $\text{\hspace{0.17em}}C=0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}D=0\text{\hspace{0.17em}}$ and work with a simplified form of the equations in the following examples.

Given the function $\text{\hspace{0.17em}}y=A\mathrm{sin}\left(Bx\right),\text{\hspace{0.17em}}$ sketch its graph.

1. Identify the amplitude, $\text{\hspace{0.17em}}|A|.$
2. Identify the period, $\text{\hspace{0.17em}}P=\frac{2\pi }{|B|}.$
3. Start at the origin, with the function increasing to the right if $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is positive or decreasing if $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is negative.
4. At $\text{\hspace{0.17em}}x=\frac{\pi }{2|B|}\text{\hspace{0.17em}}$ there is a local maximum for $\text{\hspace{0.17em}}A>0\text{\hspace{0.17em}}$ or a minimum for $\text{\hspace{0.17em}}A<0,\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}y=A.$
5. The curve returns to the x -axis at $\text{\hspace{0.17em}}x=\frac{\pi }{|B|}.$
6. There is a local minimum for $\text{\hspace{0.17em}}A>0\text{\hspace{0.17em}}$ (maximum for $\text{\hspace{0.17em}}A<0$ ) at $\text{\hspace{0.17em}}x=\frac{3\pi }{2|B|}\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}y=–A.$
7. The curve returns again to the x -axis at $\text{\hspace{0.17em}}x=\frac{\pi }{2|B|}.$

## Graphing a function and identifying the amplitude and period

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=-2\mathrm{sin}\left(\frac{\pi x}{2}\right).$

Let’s begin by comparing the equation to the form $\text{\hspace{0.17em}}y=A\mathrm{sin}\left(Bx\right).$

• Step 1. We can see from the equation that $\text{\hspace{0.17em}}A=-2,$ so the amplitude is 2.
$|A|=2$
• Step 2. The equation shows that $\text{\hspace{0.17em}}B=\frac{\pi }{2},\text{\hspace{0.17em}}$ so the period is
• Step 3. Because $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is negative, the graph descends as we move to the right of the origin.
• Step 4–7. The x -intercepts are at the beginning of one period, $\text{\hspace{0.17em}}x=0,\text{\hspace{0.17em}}$ the horizontal midpoints are at $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ and at the end of one period at $\text{\hspace{0.17em}}x=4.$

The quarter points include the minimum at $\text{\hspace{0.17em}}x=1\text{\hspace{0.17em}}$ and the maximum at $\text{\hspace{0.17em}}x=3.\text{\hspace{0.17em}}$ A local minimum will occur 2 units below the midline, at $\text{\hspace{0.17em}}x=1,\text{\hspace{0.17em}}$ and a local maximum will occur at 2 units above the midline, at $\text{\hspace{0.17em}}x=3.\text{\hspace{0.17em}}$ [link] shows the graph of the function.

Sketch a graph of $\text{\hspace{0.17em}}g\left(x\right)=-0.8\mathrm{cos}\left(2x\right).\text{\hspace{0.17em}}$ Determine the midline, amplitude, period, and phase shift.

midline: $\text{\hspace{0.17em}}y=0;\text{\hspace{0.17em}}$ amplitude: $\text{\hspace{0.17em}}|A|=0.8;\text{\hspace{0.17em}}$ period: $\text{\hspace{0.17em}}P=\frac{2\pi }{|B|}=\pi ;\text{\hspace{0.17em}}$ phase shift: $\text{\hspace{0.17em}}\frac{C}{B}=0\text{\hspace{0.17em}}$ or none

Given a sinusoidal function with a phase shift and a vertical shift, sketch its graph.

1. Express the function in the general form
2. Identify the amplitude, $\text{\hspace{0.17em}}|A|.$
3. Identify the period, $\text{\hspace{0.17em}}P=\frac{2\pi }{|B|}.$
4. Identify the phase shift, $\text{\hspace{0.17em}}\frac{C}{B}.$
5. Draw the graph of $\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{sin}\left(Bx\right)\text{\hspace{0.17em}}$ shifted to the right or left by $\text{\hspace{0.17em}}\frac{C}{B}\text{\hspace{0.17em}}$ and up or down by $\text{\hspace{0.17em}}D.$

## Graphing a transformed sinusoid

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=3\mathrm{sin}\left(\frac{\pi }{4}x-\frac{\pi }{4}\right).$

• Step 1. The function is already written in general form: $\text{\hspace{0.17em}}f\left(x\right)=3\mathrm{sin}\left(\frac{\pi }{4}x-\frac{\pi }{4}\right).$ This graph will have the shape of a sine function    , starting at the midline and increasing to the right.
• Step 2. $\text{\hspace{0.17em}}|A|=|3|=3.\text{\hspace{0.17em}}$ The amplitude is 3.
• Step 3. Since $\text{\hspace{0.17em}}|B|=|\frac{\pi }{4}|=\frac{\pi }{4},\text{\hspace{0.17em}}$ we determine the period as follows.
$P=\frac{2\pi }{|B|}=\frac{2\pi }{\frac{\pi }{4}}=2\pi \cdot \frac{4}{\pi }=8$

The period is 8.

• Step 4. Since $\text{\hspace{0.17em}}C=\frac{\pi }{4},\text{\hspace{0.17em}}$ the phase shift is
$\frac{C}{B}=\frac{\frac{\pi }{4}}{\frac{\pi }{4}}=1.$

The phase shift is 1 unit.

• Step 5. [link] shows the graph of the function.

Draw a graph of $\text{\hspace{0.17em}}g\left(x\right)=-2\mathrm{cos}\left(\frac{\pi }{3}x+\frac{\pi }{6}\right).\text{\hspace{0.17em}}$ Determine the midline, amplitude, period, and phase shift.

midline: $\text{\hspace{0.17em}}y=0;\text{\hspace{0.17em}}$ amplitude: $\text{\hspace{0.17em}}|A|=2;\text{\hspace{0.17em}}$ period: $\text{\hspace{0.17em}}P=\frac{2\pi }{|B|}=6;\text{\hspace{0.17em}}$ phase shift: $\text{\hspace{0.17em}}\frac{C}{B}=-\frac{1}{2}$

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text:
## Optimization Problems Part 2

In my last lesson, I introduced optimization problems and I discussed local extrema. You should check that out if you haven’t already. The next thing that I would like to discuss now is finding global maximums and minimums. The first step in finding a global maximum or minimum of a function is actually very similar to finding the local max and min values.

#### But once you know about the local maximums and minimums, how do you find the global extrema?

Finding the global extrema from the local extrema is really quite simple. And there really is only one way to find the global maximum and minimum values of a function. You just need to find a list of all possible x values where the global max or global min may occur. Then once you have created a list of all possibilities, you just plug them all into the original function.

Not the function’s derivative or the function’s second derivative. But the original function.

You will test for global extrema of f(x) using f(x), not f'(x) or f”(x). This is after you have your list of all possible locations where the global extrema could occur (which will require the use of f'(x)). But my point is that there wouldn’t be a first derivative test or a second derivative test with the global extrema like there was with finding local extrema.

## So what does this look like in practice?

Let’s use an example. For example, let’s say that we are asked to find the global maximum and the global minimum of $$f(x)=2x^3-\frac{5}{2}x^2+x-1$$ on the domain $$-4 \leq x \leq 5$$.

Notice we are being asked to find the global extrema on a specific domain. This is important because a lot of functions either go to infinity or to negative infinity as x either gets infinitely large, infinitely small, or approaches some specific value. So if we were asked to find the global maximum of a function that goes to infinity as x goes to infinity, we wouldn’t be able to do this. There would be no maximum since the function only continues to grow.

So we know we will be limited to a specific domain.

As I said before, finding global extrema starts out exactly like finding local extrema. The first thing we need to do is find the critical values of our function. To do this, we just need to find its derivative and set $$f'(x)=0$$ and solve for x.

$$f(x)=2x^3-\frac{5}{2}x^2+x-1$$ $$f'(x)=6x^2-5x+1$$ Then set $$6x^2-5x+1=0$$ and solve for x to find the critical values. To do this, we can factor the left side of the equation. $$(3x-1)(2x-1)=0$$ To solve this we can set each factor equal to zero individually. $$3x-1=0 \ \ \ \ and \ \ \ \ 2x-1=0$$ $$3x=1 \ \ \ \ and \ \ \ \ 2x=1$$ $$x=\frac{1}{3}, \ \frac{1}{2}$$

So now we know that this function has two critical values, $$x= \frac{1}{3}$$ and $$x=\frac{1}{2}$$. Now this is where things get different with a global max/min problem versus finding the local max/min. We also need to consider the endpoints of our given domain as critical values!

This will always be the case when we are looking for a global maximum or minimum. The problem asked us to find the global extrema on the domain $$-4 \leq x \leq 5$$. Therefore, we will also say that $$x=-4$$ and $$x=5$$ will be treated as critical values that we need to test.

### So how to we test our critical values?

The first thing that I would like to do is list out all of the x values we will be testing in one place. Remember, the list of values we need to test came from two places:

1. Setting $$f'(x)=0$$ and solving for x.
2. Each of the endpoints of the domain on which we need to find the global maximums and minimums.

So in this case we’ll have four total x values that we need to test: $$x=-4, \ \frac{1}{3}, \ \frac{1}{2}, \ and \ 5$$ To test these points, all we need to do is plug each of the four points into f(x). Whichever on outputs the largest number will tell us the global maximum. Whichever outputs the smallest number will tell us the global minimum.

$$f(-4)= \ 2(-4)^3-\frac{5}{2}(-4)^2+(-4)-1 \ = -173$$ $$f \bigg( \frac{1}{3} \bigg) = \ 2 \bigg( \frac{1}{3} \bigg) ^3-\frac{5}{2} \bigg( \frac{1}{3} \bigg) ^2+ \bigg( \frac{1}{3} \bigg) -1 \ = -\frac{47}{54}$$ $$f \bigg( \frac{1}{2} \bigg) = \ 2 \bigg( \frac{1}{2} \bigg) ^3-\frac{5}{2} \bigg( \frac{1}{2} \bigg) ^2+ \bigg( \frac{1}{2} \bigg) -1 \ = -\frac{7}{8}$$ $$f(5)= \ 2(5)^3-\frac{5}{2}(5)^2+(5)-1 \ = \frac{383}{2}$$

So we can see that the smallest of these four numbers is -173 and the largest of them is $$\frac{383}{2}$$. Therefore, the global maximum of f(x) on $$-4 \leq x \leq 5$$ is $$\frac{383}{2}$$ which occurs when $$x=5$$. And the global minimum of f(x) on $$-4 \leq x \leq 5$$ is -173 which occurs when $$x=-4$$.

We can even graph our function using Desmos along with the critical points to make sure our answer makes sense. You can click on the link in the last sentence to see a larger version of the graph.

## Extra practice

If you’d like some extra practice finding global maximums and minimums, here are a few examples you can work through on your own. A couple of these examples will require the use of the product rule and quotient rule.

For each of the following, find the global maximum and minimum of the given function on the given domain or explain why one doesn’t exist.

$$f(x)= 2x^4 + 5x^2 – 12x \ \ \textrm{on the domain} -1 \leq x \leq 2$$ $$g(x)= xe^x +6x^3-12 \ \ \textrm{on the domain} -3 \leq x \leq 0$$ $$h(x)= \frac{x^4-3x^2+1}{x+1} \ \ \textrm{on the domain} -2 \leq x \leq \frac{3}{2}$$ $$j(x)= \frac{x^4-3x^2+1}{x+1} \ \ \textrm{on the domain} -\frac{9}{10} \leq x \leq \frac{3}{2}$$

Hopefully all of this helps with global maximums and minimums, but as always I’d love to hear your questions if you have any. If you find that you get stuck as you’re working through some of these extra practice problems just let me know. Just email me at jakesmathlessons@gmail.com and I’ll see if I can help provide a bit more clarification. You can also use the form below to subscribe to my email list and I will send you my bonus FREE calculus 1 study guide to help you survive calculus! Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about derivatives as well.

## Optimization Problems

Optimization problems are another common application of the derivative. Usually in these problems you are given some function or described some situation and are then asked to find different maximums and minimums. There are a few different things that are commonly asked that you optimize, so I’d like to go over the different categories with you.

## Local maximums and minimums

The most common thing that comes up in optimization problems is finding the local maximums and minimums. The best way to do this is using the derivative of the function you are trying to optimize. Taking the function’s derivative will tell you everything you need to know.

In order to find the list of all x values where the local extrema may occur, you just need to take the function’s derivative, set it equal to zero, and solve for x. In other words, you can find the x values that will give you local max and min values by setting up the equation $$f'(x)=0$$ and solving it for x. Keep in mind, this equation often has multiple solutions, so make sure you include all possible solutions.

Doing this will give you a list of x values where all possible local maximums and minimums occur. These are called critical numbers.

Once you have your list of critical points, you will often need to figure out which ones are maximums and minimums. There are two tests you can conduct to figure which one they are.

### First derivative test

This is usually the method I like to use. As you might guess, the first derivative test only requires the use of the first derivative. I usually use this test because we already had to find the first derivative to get our list of critical values.

The first thing I would suggest doing before beginning your test is drawing out a number line and putting your critical values on it. Let’s just say for example that we had some function f(x), took its derivative, and found that the critical values are $$x=-1, \ 2, \ 6$$. Our number line might look something like this.

That’s all you need on your number line at this point. Don’t label any extra x values besides your critical values. It will only make things more confusing later.

Now all we need to do is plug x values into f’ that are around these critical values to figure out where f is increasing and decreasing. So we will need to plug in some number that is in each of the following 4 intervals. $$x<-1,$$ $$-1 < x < 2,$$ $$2 < x < 6,$$ $$x>6.$$ So all we need to do is just plug in some number in each segment of our labeled number line.

It doesn’t matter which number you plug in from each of those intervals, so you can pick whichever numbers seem easiest to plug into f’. Let’s say we will plug $$x=-2, \ 0, \ 4, \ 7$$ into f’. We want to plug them into f’ because we are trying to figure out information about the slope of f. This will tell us where it’s increasing and decreasing. Let’s imagine we plug these four x values into f’ and find that $$f'(-2)=-4, \ \ \ f'(0)=6, \ \ \ f'(4)=2, \ \ \ f'(7)=-7.$$

We only really care if these values are positive or negative. If f’ is positive at a certain x value, we know f must have a positive slope. And if f’ is negative at a certain x value, we know f must have a negative slope.

Since f'(-2) is negative, f must have a negative slope at $$x=-2$$. And f must also have a negative slope for all $$x<-1$$ since that is the interval from our number line that $$x=-2$$ falls within. So we should label this interval with a negative slope, like this:

Then we want to do the same thing for the interval of $$-1 \leq x \leq 2$$. We found out that $$f'(0)=6$$, which is a positive number. Therefore, f must have a positive slope for all $$-1 \leq x \leq 2$$. So we can label our number line accordingly.

Then we want to do the same thing with the other two intervals. This would give us something like this for our number line:

Now we just need to use this number line to determine which critical values are maximums and which are minimums. There are really only 3 main cases you need to think about for each critical value.

1. If f is increasing to the left and decreasing to the right, that critical point will be a local maximum. This will cause this little section of the graph to look like a frowny face. Therefore, the critical point will higher than the graph right around it.
2. If f is decreasing to the left and increasing to the right, that critical point will be a local minimum. This will cause this little section of the graph to look like a smiley face. Therefore, the critical point will lower than the graph right around it.
3. If f is decreasing to the left and the right or if it’s increasing to the left and the right, that critical point will NOT be a local maximum or a local minimum.

So let’s look at each of our three critical values on the number line above and see which category they all fall into.

• For $$x=-1$$, you can see that f is decreasing on the left side and increasing on the right side. Therefore, the section of the f(x) right around $$x=-1$$ looks like a smiley face and would be a local minimum.
• For $$x=2$$, you can see that f is increasing on the left side and increasing on the right side. Therefore, the critical value $$x=2$$ would not be a local maximum or a local minimum. We would need f(x) to change direction at $$x=2$$ for it to be a maximum or minimum, but that doesn’t happen here.
• For $$x=6$$, you can see that f is increasing on the left side and decreasing on the right side. Therefore, the section of the f(x) right around $$x=6$$ looks like a frowny face and would be a local maximum.

### Second derivative test

The other way you can test to see if each critical value is a local maximum or a local minimum is with the second derivative test. You do not need to use both methods if you are only trying to find local extrema because they will give you the same conclusions. Just pick which test you like more. This method will require us to find the second derivative of our function, or f”(x). We can find this simply by finding the derivative of f'(x), which we already found.

Just like with the first derivative test, it helps to draw everything out on a number line. Start with just drawing a number line that only contains the critical values which we found a while ago.

Now we need to plug each of our critical values into our second derivative, or f”(x). One important difference is that we had to plug numbers around our critical values with the first derivative test. But with the second derivative test, we will actually plug in the critical values instead of numbers around them.

Since we need to plug each critical value into our second derivative, this means we will plug $$x=-1, \ \ 2, \ \ 6$$ into f”(x). When we do that, let’s imagine we find that $$f”(-1) = 2, \ \ f”(2) = 0, \ \ f”(6) = -9.$$

Just like before, it doesn’t really matter what the exact values are that we just found. All that matters is whether they are positive, negative, or zero. If f” is positive at a certain point, then f would be concave up at that point. And if f” is negative at a point, then f is concave down at that point. If f” is zero, then f isn’t concave up or concave down at that point.

Since f”(-1) is positive (we just found that it’s 2), we know that f is concave up when $$x=-1$$. That just means that it’s curved upward, like a smiley face. So we can indicate this on our number line to keep track of what we have so far.

Since f”(2) is zero, we know that f is not concave up or down when $$x=2$$. This tells us that $$x=2$$ is the point where f switches from being concave up to concave down, or vise versa. Since f doesn’t have concavity (curvature) here, we will show this as a flat line on our number line.

And lastly, since f”(6) is negative (we just found that it’s -9), we know that f is concave down when $$x=6$$. That just means that it’s curved downward, like a frowny face. Therefore, we might get something like this.

So now we just need to figure out what all this means when it comes to the second derivative test. Again, there are three cases we want to look for.

1. If f(x) is concave up, or f”(x) is positive, for some critical value x, then this critical value represents a local minimum.
2. If f(x) is concave down, or f”(x) is negative, for some critical value x, then this critical value represents a local maximum.
3. If f(x) isn’t concave up or down, or f”(x) is zero, for some critical value x, then this critical value could be a local minimum or local maximum or neither.

So let’s compare these to our critical values to see if they are each local maximums or minimums.

• For $$x=-1$$, you can see that f is concave up. Therefore, the section of the f(x) right around $$x=-1$$ looks like a smiley face and would be a local minimum.
• For $$x=2$$, you can see that f is not concave up or concave down. In this case we don’t know from the second derivative test if this critical value would be a local maximum or a local minimum.
• For $$x=6$$, you can see that f is concave down. Therefore, the section of the f(x) right around $$x=6$$ looks like a frowny face and would be a local maximum.

Notice that these are the exact same results we found from the first derivative test, aside from the undetermined critical value. I know we didn’t actually have a function for f(x) to work through, but you would find the same thing if you did actually go through these processes with some function. To find which critical values are local maximums, local minimums, or neither, you only need to do one of these two tests.

### Extra practice

Find the critical values for the following functions and determine whether each one is a local minimum, local maximum, or neither. A couple of these examples will require the use of the product rule and the quotient rule, so check those out if you need a refresher.

$$f(x)= 2x^4 + 5x^2 – 12x$$ $$g(x)= xe^x +6x^3-12$$ $$h(x)= \frac{x^4-3x^2+1}{x+1}$$

Hopefully all of this helps you gain a bit of a better understanding of local extrema, but as always I’d love to hear your questions if you have any. Go check out part 2 of my coverage on optimization problems where I go over global maximums and minimums.

## Examples of product, quotient, and chain rules

I have already discuss the product rule, quotient rule, and chain rule in previous lessons. But I wanted to show you some more complex examples that involve these rules. The reason for this is that there are times when you’ll need to use more than one of these rules in one problem. So let’s dive right into it!

## Example 1

Find the derivative of $$y \ = \ sin(x^2 \cdot ln \ x)$$.

At first glance of this problem, the first thing we should notice is that we can think of this function as one function plugged into another. There is a clear inner function and a clear outer function.

It can be broken down as $$x^2 \cdot ln \ x$$ being plugged into sin(x) for x. Since our function can be thought of as one function plugged into another, we will want to start out with the chain rule.

### Chain rule

The first thing we need to do to apply the chain rule is to figure out our inside function and outside function. It’s usually easier to think about the insider function first.

#### Finding f and g

To find the inside function we just need to answer the question: what function is being plugged into another?

Looking at our function you can see that we are taking $$x^2 \cdot ln \ x$$ and plugging that into another function. So we will say $$g_1(x) = x^2 \cdot ln \ x$$

Now that we have decided on the inside function, we need to find the outside function. All we need to do here is look at the original function, and replace our inside function with a single x. So we will replace $$x^2 \cdot ln \ x$$ with just x. This gives us $$f_1(x) = sin(x).$$

#### Finding f’ and g’

Now that we have found f and g, we just need to take each of their derivatives to find f’ and g’.

Finding f’ should be simple here. $$f’_1(x) = cos(x)$$

Finding g’ will be a little more tricky.

Looking at our function $$g_1(x)$$ you can see that it is actually the product of two simpler functions, $$x^2$$ and $$ln \ x$$. Therefore, we are going to have to use the product rule to find this derivative. You can kind of think of this as a smaller sub-problem within our problem, so we will come back to the chain rule after applying the product rule.

### Product rule

We need to use the product rule to find the derivative of $$g_1(x) = x^2 \cdot ln \ x.$$ The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $$f_2(x)$$ and $$g_2(x)$$.

#### Finding f and g

With the product rule it doesn’t really matter which function is f and which is g. As long as we correctly identify that our function is a product of two simpler functions, it’ll work out correctly. So we will say $$f_2(x) = x^2$$ $$g_2(x) = ln \ x.$$

#### Finding f’ and g’

Now we just need to find the derivatives of f and g. Since they are fairly simple functions, this shouldn’t be too difficult.

To find the derivative of f we just need to use the power rule. $$f’_2(x) = 2x.$$

And finding the derivative of g should be a derivative that you have memorized. Using Wolfram Alpha we can see that $$g’_2(x) = \frac{1}{x}.$$

#### Plugging into the formula

Now that we have found all the pieces we need, we can simply plug them all into the product rule formula. $$g’_1(x) = f_2(x) \cdot g’_2(x) \ + \ f’_2(x) \cdot g_2(x)$$ $$\frac{d}{dx} \Big[ x^2 \cdot ln \ x \Big] \ = \ x^2 \cdot \frac{1}{x} \ + \ 2x \cdot ln \ x$$ $$\frac{d}{dx} \Big[ x^2 \cdot ln \ x \Big] \ = \ x \ + \ 2x \cdot ln \ x$$

### Back to chain rule

Now that we know the derivative of $$x^2 \cdot ln \ x$$ we can go back up to the chain rule. We knew that $$g_1(x) = \ x^2 \cdot ln \ x$$ and by using the product we just found that $$g’_1(x) = \ x \ + \ 2x \cdot ln \ x.$$ Just as a quick reminder, we already found that $$f_1(x) = sin(x)$$ $$f’_1(x) = cos(x).$$ Now we just need to plug these four pieces into the formula for chain rule. $$h’_1(x) = \ f’_1 \big( g_1(x) \big) \cdot g’_1(x)$$ $$\frac{d}{dx} \Big[ sin \big(x^2 \cdot ln \ x \big) \Big] \ = \ cos \big( x^2 \cdot ln \ x \big) \cdot \big( x \ + \ 2x \cdot ln \ x \big)$$

And that’s it! We could factor out a like term out of one of these factors, but it wouldn’t really make the function any simpler so I won’t do that. We can also use Wolfram Alpha to check our answer. A quick note on that, Wolfram Alpha uses “log” instead of “ln” to describe a natural log.

## Example 2

Find the derivative of $$y = \frac{x \ sin(x)}{ln \ x}$$.

Looking at this function we can clearly see that we have a fraction. Therefore, we can break this function down into two simpler functions that are part of a quotient. So we can see that we will need to use quotient rule to find this derivative.

### Quotient rule

As discussed in my quotient rule lesson, when we apply the quotient rule to find a function’s derivative we need to first determine which parts of our function will be called f and g.

#### Finding f and g

With the quotient rule, it’s fairly straight forward to determine which part of our function will be f and which part will be g. We will always say f is the numerator (top of our fraction) and g is the denominator (bottom of our fraction). So we can say $$f_1(x) \ = \ x \ sin(x)$$ $$g_1(x) \ = ln \ x.$$

#### Finding f’ and g’

Once we have determined which part of our function we are going to call f and which part will be g, we need to take each of their derivatives so we can use the quotient rule formula.

First we’ll start with finding f’. To find this we need to find the derivative of $$f_1(x)= x \ sin(x)$$. Notice this function is actually a product of two simpler functions. So in order to find $$\mathbf{f’_1(x)}$$ we will actually need to use the product rule. This will create a smaller sub-problem for us so we will need to come back to the quotient rule in a moment.

### Product rule

As we did in the previous example, or in my product rule lesson, we need to start by determining which piece of the function $$f_1(x) = x \ sin(x)$$ will be $$f_2$$ and which will be $$g_2$$.

#### Finding f and g

When using the product rule it doesn’t really matter which piece of the product is called f and g. So we will say $$f_2(x) = x$$ $$g_2(x) = sin(x).$$

#### Finding f’ and g’

In order to use the product rule formula, we need to find the derivative of each of these pieces now. Fortunately, both of these pieces are simple functions to differentiate. $$f’_2(x) = 1$$ $$g’_2(x) = cos(x)$$

#### Plugging into the formula

Now we just need to plug the four pieces we’ve found into the product rule formula. $$f’_1(x) = \ f_2(x) \cdot g’_2(x) \ + \ f’_2(x) \cdot g_2(x)$$ $$\frac{d}{dx} \Big[ x \ sin(x) \Big] \ = \ x \cdot cos(x) \ + \ 1 \cdot sin(x)$$ $$\frac{d}{dx} \Big[ x \ sin(x) \Big] \ = \ x \ cos(x) \ + \ sin(x)$$

### Back to the quotient rule

Now that we have used the product rule to find $$f’_1(x) = \ x \ cos(x) \ + \ sin(x)$$ we need to find $$g’_1(x)$$ so we can use the quotient rule formula. Remember $$g_1(x) = ln \ x$$, which is a function whose derivative you should memorize. $$g’_1(x) = \frac{1}{x}.$$ So now we know all of the pieces we need to apply the quotient rule formula.

$$h'(x) \ = \ \frac{f’_1(x) \cdot g_1(x) \ – \ f_1(x) \cdot g’_1(x)}{g^2_1(x)}$$ $$\frac{d}{dx} \Bigg[ \frac{x \ sin(x)}{ln \ x} \Bigg] \ = \ \frac{ \Big[ \big( x \ cos(x) + sin(x) \big) \cdot ln \ x \Big] \ – \ \Big[ x \ sin(x) \cdot \frac{1}{x} \Big]}{\big( ln \ x \big)^2}$$

And now we can just simplify by distributing through all of our parenthesis.

$$\frac{d}{dx} \Bigg[ \frac{x \ sin(x)}{ln \ x} \Bigg] \ = \ \frac{ x \ ln(x) \ cos(x) \ + \ ln(x) \ sin(x) \ – \ sin(x)}{ln^2(x)}$$

And that’s the answer! Again, we can check this using Wolfram Alpha.

If you have any questions on any of this just let me know! You can email me at jakesmathlessons@gmail.com. You can also use the contact form below and I’ll add you to my email list and send you my calculus 1 study guide to help you boost your calculus scores! I’d also love to hear any suggestions for future posts so please don’t hesitate to reach out to me. If you want some more practice with derivatives go check out my other lessons and problems related to derivatives.

## Solution – A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?

A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?

### 1. Draw a sketch

Here we have a related rates problem.  As I said when I discussed related rates problems initially, the first thing I like to do with these problems is draw a sketch of the scene that is being described.  If you want to refer back to that, you can click here.  Otherwise, let’s sketch the problem described here.

### 2. Come up with your equation

The next thing we need to do is set up our equation which will relate our different quantities.  To do this, we will want to consider what value the question is asking us to find.

What are we looking for?

It asks “at what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?”  Therefore, the value we are looking for is the “rate of change of the angle between the string and the horizontal.”  This just means we will need to consider the angle between the string and the ground (the ground is the horizontal in this case).  If you look back at our drawing, you will see that this angle is represented by $$\theta$$.  Since our goal is to find how fast $$\theta$$ is changing, we need $$\theta$$ to be in our original equation.

Now we need to consider what other quantities or variables we know something about.  Clearly we know something about the two sides of the triangle that are labeled as being 100 ft and 200 ft.  And we can use these two sides to figure out the length of the third side, which is not labeled in our drawing.

Although we could simply call one of those sides $$a$$ and the other one $$b$$ and proceed from there, there is another option that may simplify our problem.

Consider the fact that the kite is moving horizontally.  This means that the kite is not getting any further from or closer to the ground as it moves.  Therefore, the side that is labeled 100 ft will actually be 100 ft at any point in this kite’s flight.  Because of this we actually don’t need to designate a variable to this side of the triangle.  Instead this side is simply a constant 100 ft.

Now we just need to use one of the other two sides of the triangle.  We could technically use either one, but one will be a lot easier than the other.  It looks like the hypotenuse would be the easier of the two, because we know it’s 200 ft at this moment.  However, we don’t know exactly how fast it’s changing.  We can figure that out but it wouldn’t be easy.

We do know exactly how fast the unlabeled side is changing.  The question states that the kite is moving horizontally at a speed of 8 $$\frac{ft}{s}$$.  Since this unlabeled side is exactly horizontal, we know its rate of change is also 8 $$\frac{ft}{s}$$.  We can figure out its length using Pythagorean Theorem later, but this would certainly be easier than finding the rate of change of the hypotenuse.  Therefore, I will go ahead and use the unlabeled side.

Since this unlabeled side is going to be changing we will need to designate a variable to this side of the triangle.  As the kite moves away from the person flying it, the person holding the string has to let more string out and allow it to become longer.  This means that this unlabeled side in our drawing will need to be described with a variable.  We will call it side $$a$$.

Putting it into an equation.

Now we have three different quantities we need to relate somehow:

1. Angle $$\theta$$ (this will be changing as the kite moves).
2. Side $$a$$ (this will be changing as the kite moves and the string is let out).
3. Side labeled 100 ft (this will not change and can be treated as a constant).

So we have two sides and an angle that we need to make an equation with.  To do this, think about where these sides are in relation to the angle $$\theta$$.  The side labeled 100 ft is the side opposite to the angle $$\theta$$ and the side we’re calling $$a$$ is adjacent to the angel $$\theta$$.

Usually when dealing with two sides and one angle of a triangle, you will want to use either sine, cosine, or tangent to relate the three.  So which one should be used when we know the opposite side and the adjacent side to the angle in question?

Remember soh, cah, toa?

• Sine Opposite Hypotenuse

Since we have the opposite side and the adjacent side, we want to use tangent.  Therefore we can say:

$$tan(\theta) = \frac{100}{a}$$

Since it will make finding the derivative easier, I am going to rewrite this as

$$tan(\theta) =100a^{-1}$$

### 3. Implicit differentiation

As with any related rates problem, we now need to take the derivative of both sides of the equation with respect to time.  Since $$\theta$$ and $$a$$ are both functions of time, we will need to use chain rule for both sides of this equation.  We know they are functions of time because they are both going to be dependent on the position of the kite as time progresses.  We don’t have an explicit formula for either of these functions, but we know their values are dependent on time.

$$\frac{d}{dt}tan(\theta) =\frac{d}{dt}100a^{-1}$$

$$\frac{d}{dt}\frac{sin(\theta)}{cos(\theta)} =\frac{d}{dt}100a^{-1}$$

To find the derivative of the left side of this equation you will need to use the quotient rule and the chain rule.  I’m not going to show all the steps of how to do this but if you want a refresher, you can read about the quotient rule here and the chain rule here.  Using Wolfram Alpha, you can see that

$$\frac{d}{dx}tan(x)=\frac{1}{cos^2x}$$

Therefore, we can say that

$$\frac{d}{dt}tan(\theta)=\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt}$$

Plugging this back into the left side of our equation, we get

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =\frac{d}{dt}100a^{-1}$$

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt}$$

### 4. Solve for desired rate of change

The last step of any related rates problem is to solve for the desired rate of change.  Now remember the thing we need to find is the rate of change of our angle $$\theta$$.  This is exactly what $$\frac{d\theta}{dt}$$ represents.  So now we just need to solve for $$\frac{d\theta}{dt}$$.

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt}$$

$$\frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt} \cdot cos^2 \theta$$

Now we just need to plug in the values for $$a$$, $$\frac{da}{dt}$$, and $$\theta$$ and we will have our answer.  We don’t know all of these values but we can find them.

Finding a

As I mentioned before, we can find $$a$$ by using Pythagorean Theorem.  Looking back at our drawing, we have a right triangle with side lengths of 100 ft, 200 ft, and $$a$$.  We know that

$$100^2 + a^2 = 200^2$$

$$10,000 + a^2 = 40,000$$

$$a^2 = 30,000$$

$$a = \sqrt{30,000}$$

$$a = 100\sqrt{3}$$

Finding  $$\mathbf{\frac{da}{dt}}$$

This was actually given.  We know that $$a$$ is the horizontal distance the kite is away from the person flying the kite.  We know that the kite is moving horizontally at a speed of 8 $$\frac{ft}{s}$$.  Because of this we know that this is also the rate at which $$a$$ is changing.  Since $$\frac{da}{dt}$$ is the rate of change of $$a$$, we know

$$\frac{da}{dt} = 8$$

Finding $$\mathbf{\theta}$$

To find $$\theta$$ we will need to go back to the original equation we came up with before the implicit differentiation step:

$$tan(\theta) = \frac{100}{a}$$

Since we know $$a$$, we can plug it in here and solve for $$\theta$$.

$$tan(\theta) = \frac{100}{100\sqrt{3}}$$

$$tan(\theta) = \frac{1}{\sqrt{3}}$$

This angle is actually on the unit circle and by using this we know:

$$\theta = \frac{\pi}{6}$$

Note that $$\theta$$ will be in radians.

Now we can plug all of these into our equation for $$\frac{d\theta}{dt}$$.

$$\frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt} \cdot cos^2 \theta$$

$$\frac{d\theta}{dt} =-100 \big(100\sqrt{3} \big)^{-2} \cdot 8 \cdot cos^2 \Bigg( \frac{\pi}{6} \Bigg)$$

$$\frac{d\theta}{dt} =-\frac{1}{300} \cdot 8 \cdot \Bigg( \frac{\sqrt{3}}{2} \Bigg)^2$$

$$\frac{d\theta}{dt} =-\frac{1}{300} \cdot 8 \cdot \frac{3}{4}$$

$$\frac{d\theta}{dt} =-\frac{1}{50}$$

So we can say that the angle between the string and the horizontal is decreasing at a rate of $$\frac{1}{50} \ \frac{radians}{s}$$ when 200 ft of string has been let out.

And that’s the answer to the question!  Hopefully that wasn’t too bad, but if you have any questions I’d love to hear them.  I know related rates problems can be challenging so you can email me any questions or suggestions at jakesmathlessons@gmail.com.  If you have any other problems you’d like to see worked out go ahead and send me an email.

If you feel you need some more practice with related rates, you can check out the lesson where I discussed related rates for more examples.

Also, if you want to check out some other problems and get some practice with derivatives, go check out my derivatives page.  You can see what other topics I’ve already covered and problems I’ve worked through.  If you can’t find your problem there just let me know and I may post the solution to your problem.

## The Chain Rule

The chain rule is another trick for taking complex derivatives by breaking them down into simpler parts.  Rather than using this when we are multiplying or dividing two functions, we use the chain rule when our complex function can be thought of as plugging one function into another one.  These are known as composite functions.

Let’s say we have a function called $$h(x)$$ which is the composition of two simpler functions, $$f(x)$$ and $$g(x)$$ where: $$h(x)=f(g(x)).$$

Then, $$h'(x)=f'(g(x))\cdot g'(x)$$

This is known as the chain rule.

The phrase I use to remember The Chain Rule is:

The derivative of the outside, leave the inside function alone.  Then multiply by the derivative of the inside.

Similarly to the product rule and quotient rule, the first thing you need to do after identifying you need to use the chain rule is figure out which part of the function you want to call $$f(x)$$ and what to call $$g(x)$$.  Once you figure out which part to call $$f(x)$$ and $$g(x)$$, the rest of the process is almost identical to applying the product and quotient rules.

I would like to show a few examples of how to assign $$f(x)$$ and $$g(x)$$ then we can go through one of them all the way to the end.

## Example 1

Find the derivative of $$h(x)=\sqrt{x^3-4x^2+7x+1}$$.

#### Recognizing when to use the chain rule

The way I like to think about breaking it down into $$f(x)$$ and $$g(x)$$ is to consider which is the outside function and the inside function.  In this case, it is pretty clear that we have $$x^3-4x^2+7x+1$$ all inside of a square root.  Therefore, we can think of the square root as the outside function and the $$x^3-4x^2+7x+1$$ as the inside function.  In other words, we are plugging our inside function into our outside function.

Since we can say that the above function, $$h(x)$$, can be described by one function being plugged into another function, this tells us that we can use the chain rule to find its derivative.

#### Determining f and g

As mentioned before, the first thing you need to do is isolate the inside and outside functions.  I think that it is usually easier to decide on the inside function first.  The reason for this is that the inside function is often surrounded by parenthesis, or in this case, a square root.

So as a result, we can say that our inside function is $$g(x)=x^3-4x^2+7x+1.$$

Once we have figured out our inside function, we need to write everything else that’s left over as an isolated function, which we will call $$f(x)$$.  The simplest way to do this is look at our original function, $$h(x)$$, and replace the entire inside function, $$g(x)$$, with a single $$x$$.

Since $$h(x)=\sqrt{x^3-4x^2+7x+1}$$, all we need to do is replace the entire part which we have called the inside function, which is $$x^3-4x^2+7x+1$$, with a single $$x$$.  The function we are left with after doing this will be our outside function, which will be called $$f(x)$$.

Doing this leaves us with:

$$f(x)=\sqrt{x}$$

Now that we have figured out $$f(x)$$ and $$g(x)$$, we just need to figure out $$f'(x)$$ and $$g'(x)$$ and plug these functions into the chain rule formula as shown above.

I will work one of these all the way to the end a little later, but for now I’d like to do a couple more examples up to this point.

## Example 2

Find the derivative of $$h(x)=2sin(x^2+4)$$.

#### Determining f and g

Like last time, the first thing we need to do is determine what we will call our inside function and our outside function.  First we will figure out the inside function.  As I said before, the inside function is usually a bit easier to see because an easy place to start is to simply take the part of the function inside parenthesis.

If we do this in this case, we would say that our inside function is $$g(x)=x^2+4$$.  I would like to point out that this will not always work every time.  However, it is a good place to start.  You can always come back to this step if you realize your original choice for the inside function doesn’t work well.

If we say that our inside function is the $$x^2+4$$ part, all we need to do to figure out the outside function is replace that piece with a single $$x$$ and see what we have left.  If we do this, we are left with $$f(x)=2sin(x)$$.

Now we have two functions, $$f(x)=2sin(x)$$ and $$g(x)=x^2+4$$, that are much easier to derive.  Therefore, we can use these functions, find their derivatives, and put all those pieces into the chain rule formula.

Now I would like to do one example all the way from beginning to end.

## Example 3

Find the derivative of $$h(x)=\big(x^3+18\big)^{56}$$.

#### Determining f and g

Just like the first two examples, the first thing we want to do is determine our inside and our outside functions.  As before, let’s start with the inside function.  This is another case where our inside function is fairly obvious because it’s surrounded by a set of parenthesis.  Therefore, we will call our inside function $$g(x)=x^3+18$$.

Once we have our inside function, we need to determine our outside function.  To do this, we will go back to our original function and replace the inside function with a single $$x$$.  So we will replace the $$x^3+18$$ all with a single $$x$$.  Doing this gives us $$f(x)=x^{56}$$.

#### Dealing with each piece of the formula

Now we have determined our outside and inside functions to be $$f(x)=x^{56}$$ and $$g(x)=x^3+18$$.  Now, in order to use the chain rule formula to find the derivative of our original function $$h(x)$$ we also need to find $$f'(x)$$ and $$g'(x)$$.

Both of these derivatives can be found simply using the power rule.  If you can’t remember how to do this, I discussed the power rule in this article here.  All you need to do is move the power down in front of the $$x$$ and lower its power by $$1$$.

Doing this gives us:

$$f'(x)=56x^{55}$$

$$g'(x)=3x^2$$

#### Putting it all together

Now, similarly to the product and quotient rule, once we have the four necessary components we can just plug them into the chain rule formula and simplify.  Remember,

$$h'(x)=f'(g(x))\cdot g'(x)$$

Before I proceed I would like to quickly explain what the notation $$f'(g(x))$$ means.  Think about what it means to find, for example, $$f(2)$$.  All this means is that you need to plug $$2$$ into your function called $$f$$.  Or in other words, go to your function $$f(x)$$ and replace every $$x$$ with a $$2$$.  By this same reasoning, $$f'(g(x))$$ means we need to take our function $$f’$$ and replace each $$x$$ with our entire function $$g(x)$$.  So we will change each $$x$$ in our equation into $$(x^3+18)$$.  I suggest putting parenthesis around the entire function when you plug it into each $$x$$ because this will help make sure you simplify correctly.

Now let’s go ahead and use the formula.  We will plug $$g(x)$$ into $$f'(x)$$, then multiply that whole piece by our entire function $$g'(x)$$.

$$h'(x)=\Big(56\big(x^3+18\big)^{55}\Big)\Big(3x^2\Big)$$

Now you just want to simplify, and this tells you that our final derivative is:

$$h'(x)=168x^2\big(x^3+18\big)^{55}$$

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Go check out my other lessons on the derivatives page.  There’s a lot of topics covered there that are worth taking a look at.  You can also see some more practice problems using the chain rule here.  And don’t forget, if you have any questions about this article or any suggestions for future lessons I haven’t touched on, leave a comment or email me at jakesmathlessons@gmail.com!

## The Quotient Rule

Similar to the product rule, the quotient rule is a tool for finding complex derivatives by breaking them down into simpler pieces.  There is one main difference between the two.  As the names imply, the product rule is applicable when you need to find the derivative of a function that is actually the product of two simpler functions and the quotient rule is used when your function can be described as one simpler function being divided by another.

Like the product rule, there are a few different ways you might see the quotient rule represented.  I recommend picking the one that makes the most sense to you so that you can memorize the formula.

For our functions $$f(x)$$ and $$g(x)$$:

$$\frac{d}{dx}\Bigg[\frac{f(x)}{g(x)}\Bigg]=\frac{\frac{d}{dx}\big[f(x)\big]g(x)-f(x)\frac{d}{dx}\big[g(x)\big]}{\big(g(x)\big)^{2}}$$

$$\Bigg(\frac{f}{g}\Bigg)’=\frac{f’\cdot g-f\cdot g’}{g^2}$$

$$\Bigg(\frac{f(x)}{g(x)}\Bigg)’=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$$

Unlike the product rule, the order does matter here.  This is because we are using subtraction and division rather than addition and multiplication.  Since we need the pieces to be in the correct order, it is helpful to come up with some method for memorizing the quotient rule as I have shown it above.

## How to remember the quotient rule

We need to take the derivative of some function, that can be represented as a fraction made up of two functions that are easier to derive.  In other words we will be considering the “top” of the fraction and the “bottom” of the fraction as two separate, simpler functions.  The method I like to use to remember this formula is to think of the function in the numerator and the “high” portion and the denominator as the “low” portion.  These will be abbreviated as “Hi” and “Lo.”

$$\Bigg(\frac{f(x)}{g(x)}\Bigg)’=\Bigg(\frac{Hi}{Lo}\Bigg)’=\frac{Lo \ dHi-Hi \ dLo}{Lo^2}$$

And if you read it out loud, it almost seems to have a little jingle to it, which makes is easier to remember:

Lo d Hi minus Hi d Lo, all over Lo squared.

Just to add a little clarification, the “d” means you should take the derivative of the following piece.  So, “d Hi” should be where you plug in the derivative of the function that makes up the top half of the fraction.

## Example 1

Find the derivative of $$h(x)=\frac{e^x}{\sqrt{x}}$$.

#### Recognizing when to use the quotient rule

Just like we did with the product rule example, we want to first recognize how we will split up this function.  In other words, we need to recognize which part we will consider $$f(x)$$ and which part we will consider $$g(x)$$.  The main difference is that this distinction does matter with the quotient rule.  We need to call $$f(x)$$ our top function and $$g(x)$$ our bottom function.

Therefore, we need to say $$f(x)=e^x$$ and $$g(x)=\sqrt{x}$$.  Once we have made this distinction, we can consider these two functions individually for a moment and find each of their derivatives.  I already found these derivatives in Example 1 of The Product Rule.  If you want to see this again, click here.

By using the previous example, we already know $$f'(x)$$ and $$g'(x)$$.  I recommend writing out $$f(x)$$, $$f'(x)$$, $$g(x)$$, and $$g'(x)$$ all in one place before using the quotient rule formula.  So far we have:

$$f(x)=e^x$$

$$g(x)=\sqrt{x}=x^\frac{1}{2}$$

$$f'(x)=e^x$$

$$g'(x)=\frac{1}{2}x^{-\frac{1}{2}}$$

#### Putting it all together

Now that we have figured out these four parts, we can simply plug them into the quotient rule formula we have above.  This tells us that:

$$h'(x)=\Bigg(\frac{e^x}{\sqrt{x}}\Bigg)’=\frac{\Big(x^\frac{1}{2}\Big)(e^x)-(e^x)\Big(\frac{1}{2}x^{-\frac{1}{2}}\Big)}{\big(\sqrt{x}\big)^2}=\frac{e^x\Big(x^\frac{1}{2}-\frac{1}{2}x^{-\frac{1}{2}}\Big)}{x}$$

$$=e^x\Big(x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{3}{2}}\Big)$$

Please don’t forget to leave a comment or email me at jakesmathlessons@gmail.com with any questions or suggestions for future lessons!  Also, I recommend going to the derivatives page to see other topics I’ve covered.  And you can see more quotient rule practice problems here.  Just send me an email if you can’t find what you’re looking for and I’ll see what I can do to help!

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