MCQ
If $x^y=e^{x-y}$, then $\frac{d y}{d x}$ is
  • A
    $\frac{1+x}{1+\log x}$
  • B
    $\frac{1-\log x}{1+\log x}$
  • $\frac{\log x}{(1+\log x)^2}$
  • D
    not defined

Answer

Correct option: C.
$\frac{\log x}{(1+\log x)^2}$
(c) : We have, $x^y=e^{x-y}$
Taking $\log$ on both sides, we get
$
y \log x=(x-y) \log e \Rightarrow y=\frac{x}{1+\log x}
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{(1+\log x)-x \cdot \frac{1}{x}}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2}
$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free