Question

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

• A. $\frac{1+x}{1+\log x}$
• B. $\frac{1-\log x}{1+\log x}$
• C. not defined
• D. $\frac{\log x}{{(1+\log x)}^2}$

Answer 1

Answer: (D)

$x^y=e^{x-y}$
Taking log on both sides, we get
$y\log x=x-y$
$\Rightarrow$ $y(\log x+1)=x$
$\Rightarrow$ $y=\frac{x}{1+\log x}$
On differentiating w.r.t $x,$
we get $\Rightarrow$ $\frac{dy}{dx}=\frac{(1+\log x)-x\times \frac{1}{x}}{{(1+\log x)}^2}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{\log x}{{(1+\log x)}^2}$