Question

If $x=e^{\frac{x}{y}}$ , prove that
$\frac{dy}{dx}=\frac{x-y}{xlogx}$

Answer 1

Differentiate the above expression w.r.t. x
Given: $x=e^{\frac{x}{y}}$
Now lets prove $\frac{dy}{dx}=\frac{x-y}{x\log x}$
Differentiating both sides w.r.t. x
$\Rightarrow \frac{dx}{dx}=\frac{d}{dx}\left(e^{\frac{x}{y}}\right)$
$\Rightarrow 1=e^{\frac{x}{y}}\left[\frac{d}{dx}\left(\frac{x}{y}\right)\right]$
$\Rightarrow 1=e^{\frac{x}{y}}\left[\frac{y.1-x.\frac{dy}{dx}}{y^2}\right]$
$[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}]$
$\Rightarrow y^2=y.e^{\frac{x}{y}}-x.\frac{dy}{dx}.e^{\frac{x}{y}}$
$\Rightarrow \frac{dy}{dx}=\frac{y(e^{\frac{x}{y}}-y)}{x.e^{\frac{x}{y}}}=\frac{e^{\frac{x}{y}}-y}{\frac{x}{y}.e^{\frac{x}{y}}}$
$\Rightarrow \frac{dy}{dx}=\frac{x-y}{xlogx}$$[\because x=e^{\frac{x}{y}}\Rightarrow \log x=\frac{x}{y}]$
Hence proved