# Thread:

1. ## Cannot find my own mistake (derivation)

Hello.

So I have this:
$x^{y} = y^{x}$
which, in case $x$ and $y$ are positive, may be graphed like:

Assuming:
$x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})$
It seems plotting $x^{y} = y^{x}$ and $ln(x^{y}) = ln(y^{x})$ should be essentially the same.

However, my real problem is when I try to get $\frac{dy}{dx}$ out of those equations.

*Unfortunately first pictures showing my thorough work had to be removed because of some reader-disruptive flaws.*

Essence of derivation of equation #1:
$\frac{d}{dx}(x^{y} = y^{x})$
$y \cdot x^{y-1} \cdot \frac{dx}{dx} + x^{y} \cdot ln(x) \cdot \frac{dy}{dx} = x \cdot y^{x-1} \cdot \frac{dy}{dx} + y^{x} \cdot ln(y) \cdot \frac{dx}{dx}$
$\frac{dy}{dx} = \frac{y^{x} \cdot ln(y) - y \cdot x^{y-1}}{x^{y} \cdot ln(x) - x \cdot y^{x-1}}$

Essence of derivation of equation #2:
$\frac{d}{dx}(ln(x^{y}) = ln(y^{x}))$
$\frac{y}{x} \cdot \frac{dx}{dx} + ln(x) \cdot \frac{dy}{dx} = \frac{x}{y} \cdot \frac{dy}{dx} + ln(y) \cdot \frac{dx}{dx}$
$\frac{dy}{dx} = \frac{ln(y) - \frac{y}{x}}{ln(x) - \frac{x}{y}}$

Why is there a big difference?

2. Originally Posted by Pranas
Hello.

So I was looking for the derivative of this:

What is that very first line? This is false big time: the left hand is $yx^{y-1}$ , whereas the right one is $y^x\ln y$ ...

Tonio

But then I tried to put everything inside logarithm ln() , which, in case numbers are positive, I thought wouldn't change the answer. However it changed big time:

Can you explain, what is wrong with my thoughts?
.

3. tonio, could you please be at least a little more thorough?

$x^{y} = y^{x}$ is what i have.
$\frac{dy}{dx} = ?$ is what I need.

As long as I understand you saying $y \cdot x^{y-1} = y^{x} \cdot ln(y)$ that doesn't give me much.

4. Originally Posted by Pranas
tonio, could you please be at least a little more thorough?

$x^{y} = y^{x}$ is what i have.
$\frac{dy}{dx} = ?$ is what I need.

As long as I understand you saying $y \cdot x^{y-1} = y^{x} \cdot ln(y)$ that doesn't give me much.

Pranas, could you please be a little less sloppy? You did not say anything about having $x^y=y^x$ . It doesn't

appear anywhere in your message. It only appears the first equality $\frac{d}{dx}(x^y)=\frac{d}{x}(y^x)$ , which gives

you what I wrote you.

And don't bother in writing back asking for my help unless you first apologize.

Tonio

5. Originally Posted by tonio
Pranas, could you please be a little less sloppy? You did not say anything about having $x^y=y^x$ . It doesn't

appear anywhere in your message. It only appears the first equality $\frac{d}{dx}(x^y)=\frac{d}{x}(y^x)$ , which gives

you what I wrote you.

And don't bother in writing back asking for my help unless you first apologize.

Tonio
Yes, I apologize. I simply couldn't figure your post out at first, although it is very correct
Seems I should have written $\frac{d}{dx}(x^{y} = y^{x})$ ?

However $x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})$ so it seems like the disagreement I get
$\frac{d}{dx}(x^{y} = y^{x}) \neq \frac{d}{dx}(ln(x^{y}) = ln(y^{x}))$ prevails although it (I believe) really means the same.

6. Originally Posted by Pranas
Yes, I apologize. I simply couldn't figure your post out at first, although it is very correct
Seems I should have written $\frac{d}{dx}(x^{y} = y^{x})$ ?

However $x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})$ so it seems like the disagreement I get
$\frac{d}{dx}(x^{y} = y^{x}) \neq \frac{d}{dx}(ln(x^{y}) = ln(y^{x}))$ prevails although it (I believe) really means the same.
Well, that last statement is certainly true- $\frac{d(x^y= y^x)}{dx}\ne \frac{d ln(x^y)= ln(y^x)}{dx}$ and no one has said they are equal.

The problem goes back to your initial post: "I was looking for the derivative of this
$\frac{d x^y}{dx}= \frac{d y^x}{dx}$
which is very ambiguous: were you asking for the derivative of each side of that equation or were you asking how to arrive at that equation?

Now you say "Seems I should have written $\frac{d}{dx}(x^{y} = y^{x})$
If we let $z= x^y$, then $ln(z)= y ln(x)$ and so $\frac{1}{z}\frac{dz}{dx}= \frac{y}{x}+ ln(x)\frac{dy}{dx}$.

That is, $\frac{dz}{dx}= \frac{d x^y}{dx}= \left(\frac{y}{x}+ ln(x)\frac{dy}{dx}\right)z$
$\frac{dx^y}{dx}= yx^{y-1}+ y^x ln(x)\frac{dy}{dx}$.

Of course, that depends upon dy/dx. We cannot differentiate some function of x without knowing precisely what that function is- what y is as a function of x.

(Note that if y does NOT depend upon x, if y is a constant, then dy/dx= 0 and this just becomes the standard $\frac{dx^y}{dx}= yx^{y-1}$ from Calculus I.)

7. Indeed there were some obvious flaws in my "questionnaire", I am trying to correct most of it as the conversation evolves.

Originally Posted by HallsofIvy
Well, that last statement is certainly true- $\frac{d(x^y= y^x)}{dx}\ne \frac{d ln(x^y)= ln(y^x)}{dx}$ and no one has said they are equal...
For the sake of simplicity, let's say we're interested only in positive values of $x$ and $y$ (I've added a possible graphing below in this post).
Please be aware that I did not add $ln()$ just like that, I tried to rationally generate it as showed in this thread:
$x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})$
Maybe I am wrong, but at this moment I do not see how were the relations between $x$ and $y$ effected by that, therefore I assume I haven't changed the possible plot nor the value of $\frac{dy}{dx}$.

Originally Posted by HallsofIvy
...The problem goes back to your initial post: "I was looking for the derivative of this
$\frac{d x^y}{dx}= \frac{d y^x}{dx}$
which is very ambiguous: were you asking for the derivative of each side of that equation or were you asking how to arrive at that equation?...
I did not write what I really meant to at that point. Sorry.

Originally Posted by HallsofIvy
...Now you say "Seems I should have written $\frac{d}{dx}(x^{y} = y^{x})$
If we let $z= x^y$, then [tex]ln(z)= y ln(x) and so $\frac{1}{z}\frac{dz}{dx}= \frac{y}{x}+ ln(x)\frac{dy}{dx}$.

That is, $\frac{dz}{dx}= \frac{d x^y}{dx}= \left(\frac{y}{x}+ ln(x)\frac{dy}{dx}\right)z$
$\frac{dx^y}{dx}= yx^{y-1}+ y^x ln(x)\frac{dy}{dx}$...
Well, yes. I pretty much applied parallel method to yours on both $x^{y}$ and $y^{x}$. Then expressed $\frac{dy}{dx}$ (that would be my equation #1 in updated first message). As far as my mental calculation goes, answer seems to be identical as the one you're approaching in this part of the post.

Originally Posted by HallsofIvy
...Of course, that depends upon dy/dx. We cannot differentiate some function of x without knowing precisely what that function is- what y is as a function of x.

(Note that if y does NOT depend upon x, if y is a constant, then dy/dx= 0 and this just becomes the standard $\frac{dx^y}{dx}= yx^{y-1}$ from Calculus I.)
Indeed you're correct once more. That is a reasonable variation in a broad sense, although might be a little vulgar mapped on a plane, because I would imagine it as of having only a few points.

What I imagined as a relation between positive $x$ and $y$ in the function $x^{y} = y^{x}$ is like this:

In my language that is defined to be a simple function, only "unexpressed", because of not being represented by $y =$*operations involving only variable $x$ and constants*

P.S. We've had some confusion here, so I can politely remind, that what's still unclear for me is inequality $\frac{d(x^{y}= y^{x})}{dx}\ne \frac{d(ln(x^{y})= ln(y^{x}))}{dx}$.
Also I tried to do my best in editing the first post.
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