Question

if $(xe{)}^y=e^x$, then $\frac{dy}{dx}$ is =

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

Answer 1

Answer: (A)

$(xe{)}^y=e^x$
$\Rightarrow y(\log x+1)=x$
$\Rightarrow y=x/(\log x+1)$
$\therefore \left(\frac{dy}{dx}=\frac{logx}{(logx+1{)}^2}\right)$