Question

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

• A. $\frac{e^x}{x}$
• B. $e^x\left(\log x+\frac{1}{x}\right)$
• C. $e^x\left(x\log x+\frac{1}{x}\right)$
• D. $\frac{e^x}{\log x}$

Answer 1

The correct option is B

$e^x\left(\log x+\frac{1}{x}\right)$

Explanation for the correct option.

Find the $\frac{dy}{dx}$:

Given that, $y=e^x\log x$

Differentiate with respect to $x$ using product rule i.e;

$\frac{\mathbf{d}\left(uv\right)}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{u}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{v}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}$.

Let $u=e^x,v=\log x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& e^x·\frac{1}{x}+\log x·e^x\\ & =& e^x\left(\frac{1}{x}+\log x\right)\end{array}$

Hence, option B is correct.