Question

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

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

Answer 1

The correct option is C $\frac{\log x}{(1+\log x{)}^2}$

$x^y=e^{x-y}$
$y\log x=x-y$
$\frac{y}{x}\log x=1-\frac{y}{x}\Rightarrow \frac{y}{x}=\frac{1}{1+\log x}\Rightarrow y=\frac{x}{(1+\log x)}$
$\Rightarrow \frac{dy}{dx}=\frac{(1+\log x)\cdot 1-x\times \frac{1}{x}}{1+\log x}$
$\frac{dy}{dx}=\frac{\log x}{(1+\log x{)}^2}$