We have the equation: $$ x^y = y^x +y $$ Which defines an implicit function $y(x)$ at the point $(2,1)$. I'm asked to find the derivative at $y'(2)$.

I saw the answer in Wolfram: $$ y'(x) = \frac{y (y x^y-x y^x \ln{y})}{x (-y x^y \ln{x}+x y^x+y)} $$ Which gives $y'(2)=\frac{-1}{\ln{4}-3}$.

I don't understand how to get there. When I try to dervie, after taking $\ln$ from both sides I get: $$ y'\ln{x} + \frac{y}{x} = \frac{y^x\ln{x} + 1}{y^x+y}y' $$ $$ y'[\frac{1 - y\ln{x}}{y^x+y}] = \frac{y}{x} $$ $$ y' = \frac{y^{x+1}+y^2}{x - xy\ln{x}} $$

  • $\begingroup$ Hint: I do not suggest taking the log of both sides because it doesn't help with $y^x+y$ because the log is not linear. $\endgroup$ – Michael Burr Aug 29 '15 at 13:49

By the property of derivatives, if we find the derivative of all individually, then we can add them all and do the implicit differentiation.
$y\log{x}=\log t$
$t(\frac{dy}{dx}\log x+\frac yx)=\frac{dt}{dx}$
For $y^x=u$
$x\log y=\log u$
$u(\log y+\frac{x}{y}\frac{dy}{dx})=\frac{du}{dx}$
So $\frac{dt}{dx}=\frac{du}{dx}+\frac{dy}{dx}$
$x^y(\frac{dy}{dx}\log x+\frac yx)=y^x(\log y+\frac{x}{y}\frac{dy}{dx})+\frac{dy}{dx}$
Substituting the point
$\frac{-1}{\log4-3}=\frac{dy}{dx}$. Which is the required result.


To implicitly derive $x^y=y^x+y$ with respect to $x$, $x$ and $y$ will need to take turns behaving as constants.

For instance, to compute $$ \frac{d}{dx}x^y, $$ first treat $x$ as a constant and then treat $y$ as a constant (and remembering $\frac{dy}{dx}$ when you take a derivative with respect to $y$). Therefore, $$ \frac{d}{dx}x^y=x^y\ln(x)\frac{dy}{dx}+yx^{y-1} $$

Similarly, $$ \frac{d}{dx}(y^x+y)=xy^{x-1}\frac{dy}{dx}+\frac{dy}{dx}+y^x\ln(y). $$

Therefore, $$ x^y\ln(x)\frac{dy}{dx}+yx^{y-1}=xy^{x-1}\frac{dy}{dx}+\frac{dy}{dx}+y^x\ln(y) $$ and you can solve for $\frac{dy}{dx}$ as $$ \left(x^y\ln(x)-xy^{x-1}-1\right)\frac{dy}{dx}=y^x\ln(y)-yx^{y-1} $$ or $$\frac{dy}{dx}=\frac{y^x\ln(y)-yx^{y-1}}{x^y\ln(x)-xy^{x-1}-1}.$$

  • 1
    $\begingroup$ Last step can be cancelled by directly substituting for $y$ and $x$. $\endgroup$ – Aditya Agarwal Aug 29 '15 at 13:56
  • $\begingroup$ @AdityaAgarwal True, but the OP seemed to also want to know where WA got it's formula from. $\endgroup$ – Michael Burr Aug 29 '15 at 14:04

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