Question

If $x^y=e^{x-y}$, prove that $\frac{dy}{dx}=\frac{\log x}{(1+\log x{)}^2}$.

Answer 1

Given: $x^y=e^{x-y}$
Taking log on both the sides, $\log \left(x^y\right)=\log \left(e^{x-y}\right)$
$y\log \left(x\right)=(x-y)\log \left(e\right)=(x-y)$ ....$\left[\log \right(e)=1]$

$y=\frac{x}{1+\log \left(x\right)}$

Differentiating w.r.t. x both the sides,
$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{x}{1+\log \left(x\right)}\right)$

$\frac{dy}{dx}=\frac{(1+\log x)\frac{d\left(x\right)}{dx}-x.\frac{d(1+\log (x\left)\right)}{dx})}{(1+\log (x){)}^2}$

$\frac{dy}{dx}=\frac{(1+\log x)\left(1\right))-x.(0+\frac{1}{x})}{(1+\log (x){)}^2}$

$\frac{dy}{dx}=\frac{\log \left(x\right)}{(1+\log (x){)}^2}$
Hence proved.