Question

If $x^y=e^{x-y}$, then $\frac{d y}{d x}$ is equal to

(A) $\frac{1}{1+\log x}$

(B) $\frac{\log x}{(1+\log x)^2}$

(C) $\frac{x}{1+(\log x)^2}$

(D) $\frac{\log x}{1+\log x}$

Answer 1

We have, $x^y=e^{x-y}$

$\Rightarrow \quad \begin{aligned}y \log x & =(x-y) \log _e e=(x-y) \\y & =\frac{x}{1+\log x}\end{aligned}$

$\begin{aligned}\Rightarrow \frac{d y}{d x} & =\frac{1+\log x-x \times \frac{1}{x}}{(1+\log x)^2} \\& =\frac{1+\log x-1}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2}\end{aligned}$

So, The correct option will be (B).