Question

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

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

Answer 1

Answer: (C)

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

$x^y=e^{x-y}$

Take ${\log }_e$ on both sidees

${\log }_e{x}^y={\log }_e{e}^{x-y}$ $\Rightarrow$ $y\log x=(x-y){\log }_{e}e$

$y\log x=x-y$ $⟶\left(1\right)$

$y[\log x+1]=x$

Differentiate both sides w.r.t. $x$

$\frac{dy}{dx}[1+\log x]+y\left[\frac{1}{x}\right]=1$

$\frac{dy}{dx}=\frac{1-\frac{y}{x}}{1+\log x}$

From eqn.(1) $\frac{y}{x}=\frac{1}{(1+\log x)}$

$\frac{dy}{dx}=\frac{1-\frac{1}{1+\log x}}{1+\log x}=\frac{\log x}{{(1+\log x)}^2}$