Question

If $x^y=e^{x-y}$, then find $\frac{dy}{dx}$ at $x=1$

Answer 1

Applying log on both sides

$ylogx=x-y$.......(1) $[\because \log x^m=m\log x]$
$y\log x+y=x\Rightarrow y=\frac{x}{1+\log x}$
When $x=1$ then $y=\frac{1}{1+\log 1}$
$=\frac{1}{1+0}$
$\therefore y=1$

Differentiating equation (1) wrt x

$\frac{y}{x}+logx\frac{dy}{dx}=1-\frac{dy}{dx}$ $[\because (UV{)}^{\prime }=U{V}^{\prime }+U^{\prime }V]$
$\log x\frac{dy}{dx}+\frac{dy}{dx}=1-\frac{y}{x}$
$\frac{dy}{dx}(\log x+1)=\frac{x-y}{x}$

$\frac{dy}{dx}=\frac{x-y}{x(1+logx)}$
To find $\frac{dy}{dx}$ at $x=1$:
$[\frac{dy}{dx}{]}_{x=1}=\frac{1-1}{1(1+\log 1)}=\frac{0}{1(1+0)}=\frac{0}{1}=0$