Question

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

Answer 1

Differentiate the above expression w.r.t. x
Given : $y^x=e^{y-x}.....\left(1\right)$
Let's prove $\frac{dy}{dx}=\frac{(1+logy{)}^2}{\log y}$
Taking log both sides, We get,
$\log y^x=\log e^{y-x}$
$\Rightarrow x\log y=y-x..........\left(2\right)$
Differentiating both sides w.r.t. x
We get,
$\frac{d}{dx}\left(xlogy\right)=\frac{d}{dx}(y-x)$
$\Rightarrow logy+\frac{x}{y}.\frac{dy}{dx}=\frac{dy}{dx}-1$
$[\because \frac{d\left(uv\right)}{dy}=v\frac{du}{dy}+u\frac{dv}{dy}]$
$\Rightarrow \frac{dy}{dx}=\frac{1+logy}{1-\frac{x}{y}}$
$Using\left(2\right)x=\frac{y}{1+logy}$
$\Rightarrow \frac{dy}{dx}=\frac{(1+logy)}{1-\frac{y}{y(1+logy)}}$
$\Rightarrow \frac{dy}{dx}=\frac{(1+logy{)}^2}{logy}$