Question

If $x^y=e^{x-y}$ then find $\frac{dy}{dx}$.

Answer 1

We have,

$x^y=e^{x-y}$

Taking log both side and we get,

$\log x^y=\log e^{x-y}$

$\Rightarrow y\log x=(x-y)\log e$

$\Rightarrow y\log x=x-y$

On differentiation and we get,

$y\frac{1}{x}+\log x\frac{dy}{dx}=1-\frac{dy}{dx}$

$\Rightarrow \log x\frac{dy}{dx}+\frac{dy}{dx}=1-\frac{y}{x}$

$\Rightarrow (\log x+1)\frac{dy}{dx}=\frac{x-y}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{x-y}{x(\log x+1)}$

Hence, this is the answer.