Question

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

• A. $\frac{logx}{log(x-y)}$
• B. $\frac{e^x}{x^{x-y}}$
• C. $\frac{logx}{(1+logx{)}^2}$
• D. $\frac{1}{y}-\frac{1}{x-y}$

Answer 1

Answer: (C)

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