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I need to find the derivative. given that:

$$x^y=y^x$$

defines: $$y=y(x)$$

Thank you!

BAM
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2 Answers2

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Consider $y=y(x)$ and use implicit differentiation on: $$x^{y(x)}=y(x)^x$$ Carefully computing the derivative (don't forget the chain rule) gives: $$x^y \left( \color{blue}{y'} \ln x + \frac{y}{x} \right) = y^x \left( \ln y + \frac{x\color{blue}{y'}}{y} \right)$$where I omitted the argument $x$ of $y$ and $y'$ to simplify the notation. Now solve for $\color{blue}{y'}$, this only requires some algebraic manipulations. You can check your answer here.

StackTD
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Method 1 \begin{align*} y^{x} &= x^{y} \\ x\ln y &= y\ln x \\ \ln y+\frac{xy'}{y} &= y'\ln x+\frac{y}{x} \\ \left( \frac{x}{y}-\ln x \right)y' &= \frac{y}{x}-\ln y \\ y' &= \frac{y(y-x\ln y)}{x(x-y\ln x)} \end{align*}

Method 2

Let $y=(t+1)x$, then \begin{align*} [(t+1)x]^{x} &= x^{(t+1)x} \\ (t+1)^{x}x^{x} &= x^{tx} x^{x} \\ (t+1)^{x} &= x^{tx} \\ t+1 &= x^{t} \\ x &= (t+1)^{1/t} \\ y &= (t+1)^{(t+1)/t} \\ \dot{x} &= (t+1)^{1/t-1} \times \frac{t-(t+1)\ln (t+1)}{t^{2}} \\ \dot{y} &= (t+1)^{1/t+1} \times \frac{t-\ln (t+1)}{t^{2}} \\ \frac{dy}{dx} &= (t+1)^{2} \frac{t-\ln (t+1)}{t-(t+1)\ln (t+1)} \\ \end{align*}

Further points:

We omit the trivial case $y(x)\equiv x$ here.

Ng Chung Tak
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    Actually, we can express it in terms of Lambert W function, $y(x)=-\frac{x}{\ln x} W\left( -\frac{\ln x}{x} \right)$, but that's also weird somehow. – Ng Chung Tak Jun 14 '16 at 15:37