Question

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

• A. $e$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (C)

Given have, $x^y={\theta }^{x-y}$
taking logon both sides, we get
$y\log x=(x-y)\log \theta =(x-y)...(i)$
when $x=1$, then $y(\log 1)=(1-y)$
$\Rightarrow y=1$
On differentiating both sides,
$y\left(\frac{1}{x}\right)+\log x\cdot \frac{dy}{dx}=1-\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}(\log x+1)=1-\frac{y}{x}$
$\Rightarrow \frac{dy}{dx}(\log x+1)=\frac{x-y}{x}$
$\Rightarrow \frac{dy}{dx}=\frac{(x-y)}{x(\log x+1)}$
when $x=1$, then
$\left(\frac{dy}{dx}\right)=\frac{1-1}{x(\log 1+1)}=0$