# Cannot find my own mistake (derivation)

Printable View

• Nov 21st 2010, 10:22 AM
Pranas
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:
http://img44.imageshack.us/img44/8409/88706802.jpg

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?
• Nov 21st 2010, 03:23 PM
tonio
Quote:

Originally Posted by Pranas
Hello.

So I was looking for the derivative of this:
http://www.ipix.lt/images/71864780.jpg

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:
http://www.ipix.lt/images/93752048.jpg

Can you explain, what is wrong with my thoughts?

.
• Nov 22nd 2010, 04:49 AM
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.
• Nov 22nd 2010, 06:04 AM
tonio
Quote:

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
• Nov 22nd 2010, 06:35 AM
Pranas
Quote:

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 :D
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.
• Nov 22nd 2010, 06:50 AM
HallsofIvy
Quote:

Originally Posted by Pranas
Yes, I apologize. I simply couldn't figure your post out at first, although it is very correct :D
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.)
• Nov 22nd 2010, 09:04 AM
Pranas
Indeed there were some obvious flaws in my "questionnaire", I am trying to correct most of it as the conversation evolves.

Quote:

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 $\frac{dy}{dx}$.

Quote:

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.

Quote:

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.

Quote:

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:
http://img44.imageshack.us/img44/8409/88706802.jpg
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.