Question

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

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

Answer 1

Answer: (A)

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