# Trying to derive an inverse trigonometric function

I'd like to know how to derive these functions (I know the answers, I want to know how to get there) \begin{align*} f(x) &= \arcsin\left(\frac{x}{3}\right)\\ f(x) &= \arccos(2x+1)\\ f(x) &= \arctan(x^2)\\ f(x) &= \mathrm{arcsec}(x^7)\\ \end{align*} etc.

-
You mean differentiate them? Use the chain rule. – Qiaochu Yuan Feb 21 '11 at 19:38

1
Are you asking how to get, for example, the derivative of the arcsine function in general? Or are you asking how to get the derivative of, for example, $f(x) = \arcsin\left(\frac{x}{3}\right)$ already knowing the derivative of the arcsine function in general? – Isaac Feb 21 '11 at 20:01
@Isaac I'm trying to get the general way of doing it, and then applying it to specific cases, like these mentioned. Thanks. – Pacane Feb 21 '11 at 20:08

1
The general way is to know the derivatives of the inverse trigonometric functions, and then apply the Chain Rule. You could try to derive the derivative of, say $f(x)=\mathrm{arcsec}(x^7)$ by finding the derivative of $\sec(f(x))$, applying the Chain Rule, and then solving for $f'(x)$, but in the long run it is probably going to be less work to remember the derivatives of the inverse trig functions than to derive them from scratch every time you need them. – Arturo Magidin Feb 21 '11 at 20:16

## 4 Answers

The formulas you need are the derivatives of $\arcsin(u)$, $\arccos(u)$, $\arctan(u)$, $\mathrm{arcsec}(u)$, and presumably $\mathrm{arccot}(u)$ and $\mathrm{arccsc}(u)$. Once you know these, you can apply the Chain Rule.

And how do you find these derivatives? Well, the Inverse Function Theorem is your first friend. If $y = g(x)$ has an inverse, is differentiable at $x=a$, $g(a)=b$, and $g'(a)\neq 0$, then $(g^{-1})'(b) = \frac{1}{g'(a)}$.

So, consider $y=\sin(\theta)$. Since the derivative of $\sin(\theta)$ is $\cos(\theta)$, you have that $$\frac{d}{du}\arcsin(u) = \frac{1}{\cos(\arcsin(u))}.$$ But... what is $\cos(\arcsin(u))$? Suppose $\arcsin(u)=\theta$. That means that $\sin(\theta) = u$, and since $\sin^2(\theta)+\cos^2(\theta)=1$, then $\cos^2(\theta) = 1 - \sin^2(\theta) = 1-u^2$. Therefore, $|\cos(\theta)|=\sqrt{\cos^2\theta} = \sqrt{1-u^2}$; and because in order to talk about the inverse of $\sin \theta$ we must have $-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2}$, then $\cos\theta\geq 0$, so $|\cos\theta|=\cos\theta$. That is, $\cos\theta = \sqrt{1-u^2}$. So, plugging into the formula for the derivative of $\arcsin(u)$, we have: $$\frac{d}{du}\arcsin(u) = \frac{1}{\cos(\arcsin u)} = \frac{1}{\sqrt{1-u^2}}.$$

Performing the same kind of analysis for $\arccos(u)$, we get $$\frac{d}{du}\arccos(u) = \frac{1}{-\sin(\arccos u)} = -\frac{1}{\sqrt{1-u^2}}.$$

For $\arctan u$, using the fact that $(\tan\theta)' = \sec^2\theta$, we have $$\frac{d}{du}\arctan u = \frac{1}{\sec^2(\arctan u)}.$$ Now, if $\arctan u = \theta$, then $\tan(\theta) = u$. Using the fact that $\tan^2\theta + 1 = \sec^2\theta$, we get that $sec^2(\arctan u) = \sec^2(\theta) = 1 + \tan^2(\theta) = 1+u^2$, so $$\frac{d}{du}\arctan u = \frac{1}{\sec^2(\arctan u)} = \frac{1}{1+u^2}.$$

For $\mathrm{arccot u}$, the same analysis works, provided you remember that $(\cot\theta)' = -\csc^2\theta$ and that $1 + \cot^2\theta = \csc^2\theta$, so $$\frac{d}{du}\mathrm{arccot}(u) = \frac{1}{-\csc^2(\mathrm{arccot}(u))} = -\frac{1}{1+u^2}.$$

With $\mathrm{arcsec}u$, we have $(sec\theta)' = sec\theta\tan\theta$, so $$\frac{d}{du}\mathrm{arcsec}(u) = \frac{1}{\sec(\mathrm{arcsec} (u))\tan(\mathrm{arcsec} u)}.$$ Here, $\sec(\mathrm{arcsec} (u)) = u$; if $\mathrm{arcsec}(u)=\theta$, then $\sec\theta = u$, and from $\tan^2\theta + 1 = \sec^2\theta$, we get $|\tan\theta| = \sqrt{u^2 - 1}$. You get: $$\frac{d}{du}\mathrm{arcsec}(u) = \frac{1}{\sec(\mathrm{arcsec}(u))\tan(\mathrm{arcsec}(u))} = \frac{1}{u\sqrt{u^2-1}}.$$

And finally, using the fact that $(\csc\theta)' = -\csc\theta\cot\theta$, you get $$\frac{d}{du}\mathrm{arccsc}(u) = \frac{1}{-\csc(\mathrm{arccsc}(u))\cot(\mathrm{arccsc}(u))} = -\frac{1}{u\sqrt{u^2-1}}.$$

Once you have these formulas, the Chain Rule takes care of the rest.

So you have: \begin{align*} \frac{d}{du}\arcsin(u) &= \frac{1}{\sqrt{1-u^2}}, &\qquad \frac{d}{du}\arccos u &= -\frac{1}{\sqrt{1-u^2}},\\ \frac{d}{du}\arctan(u) &=\frac{1}{1+u^2}, &\frac{d}{du}\mathrm{arccot}(u) &= -\frac{1}{1+u^2},\\ \frac{d}{du}\mathrm{arcsec}(u) &=\frac{1}{u\sqrt{u^2-1}}, &\frac{d}{du}\mathrm{arccsc}(u) &= - \frac{1}{u\sqrt{u^2-1}}. \end{align*}

-
Thanks for your very descriptive answer. One last thing though, I think I am not using the chain rule correctly. Let's say I take the example of $f(x) = \arcsin\frac{x}{3}$ , now I know that $\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}$ and I know that $\frac{d}{dx} \frac{x}{3} = \frac{1}{3}$ I was going to multiply both, but apparently it doesn't work like that. Got any tip for me ? – Pacane Feb 21 '11 at 20:38

2
@Pacane: Looks like your real problem is that you don't know how to apply the Chain Rule. The Chain Rule says: $$\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$. So you need to evaluate the derivative of $\arcsin(u)$ at $u=\frac{x}{3}$, then multiply by the derivative of $\frac{x}{3}$. That is:$$\frac{d}{dx}\arcsin\frac{x}{3} = \left(\frac{1}{\sqrt{1 -\left(\frac{x}{3}\right)^2}}\right)\left(\frac{x}{3}\right)' = \left(\frac{1}{\sqrt{1 - \frac{x^2}{9}}}\right)\frac{1}{3}. – Arturo Magidin Feb 21 '11 at 20:46
Thanks for your help. Really appreciated. – Pacane Feb 21 '11 at 20:54

I'll walk you through a derivation (how to get there) of the derivative of the arcsine function. The same idea can be applied to the other inverse trigonometric functions. If $y=\arcsin x$, then $\sin y=\sin(\arcsin x)=x$. $$\sin y=x$$ Take the derivative of both sides with respect to $x$ (remember the chain rule!). $$\cos y\cdot\frac{dy}{dx}=1$$ Isolate $\frac{dy}{dx}$, which is what we're trying to find. $$\frac{dy}{dx}=\frac{1}{\cos y}$$ Since we want $\frac{dy}{dx}$ in terms of $x$, substitute for $y$. $$\frac{dy}{dx}=\frac{1}{\cos (\arcsin x)}$$ $\cos(\arcsin x)=\sqrt{1-x^2}$ (see this answer of mine for a technique for simplifying a trig function of an inverse trig function), so $$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}.$$

-

When you say "derive these functions" do you mean "take the derivative of these functions?" If so, the chain rule is your friend. So for $f(x)=\arctan(x^2)$, $f'(x)=\frac{1}{1+(x^2)^2}2x$

-
How did you get the $\frac{1}{1+(x^2)^2}$ part ? – Pacane Feb 21 '11 at 19:43

1
@Pacane: From $\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$. – Arturo Magidin Feb 21 '11 at 19:45

if $y=\arcsin(x)$ then $\sin(y)=x$ and differentiating wrt $x$ we get $cos(y)y'=1$. So, $y'=\frac{1}{\cos(\arcsin(x))}$. draw a triangle (with sides 1, $x$, $\sqrt{1-x^2}$) to see that $\cos(\arcsin(x))=\sqrt{1-x^2}$. Hence $\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$. This kind of reasoning works for various inverse function (using implicit differentiation). see any calc textbook under implicit differentiation.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 73, "mathjax_display_tex": 15, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8767760396003723, "perplexity_flag": "head"}