Question

If $y=x^{e^x}$, then $\frac{dy}{dx}=$

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

Answer 1

The correct option is D $y{e}^x(logx+\frac{1}{x})$

$logy=e^{x}logx$
$\therefore \frac{1}{y}\frac{dy}{dx}=e^{x}logx+e^x.\frac{1}{x}=e^x(logx+\frac{1}{x})$
$\therefore \frac{dy}{dx}=y{e}^x(logx+\frac{1}{x})$