Question

Differentiate the function w.r.t.x:
$\rm \frac{x\log x}{e^{x}}$

Answer 1

Answer:

$\frac{d}{dx} \rm \frac{x\log x}{e^{x}}= \frac{ (1 + log \: x -x \: log \: x)}{ {e}^{x} }$

Step-by-step explanation:

To differentiate the function w.r.t.x:

$\rm \frac{x\log x}{e^{x}}$

apply U/V rule of differentiation

$\frac{d}{dx} \frac{u}{v} = \frac{v. \frac{du}{dx} -u. \frac{dv}{dx} }{ {v}^{2} } \\ \\ \frac{d}{dx} \rm \frac{x\log x}{e^{x}} \: = \frac{ {e}^{x} . \frac{d(x \: log \: x)}{dx} -(x \: log \: x). \frac{d {e}^{x} }{dx} }{ {e}^{2x} } \\ \\ = \frac{ {e}^{x} . (x \frac{1}{x} + log \: x) -(x \: log \: x). {e}^{x}}{ {e}^{2x} } \\ \\ = \frac{ {e}^{x} . (1 + log \: x) -(x \: log \: x). {e}^{x}}{ {e}^{2x} } \\ \\ = \frac{ {e}^{x} . (1 + log \: x -x \: log \: x)}{ {e}^{2x} } \\ \\ \\ \frac{d}{dx} \rm \frac{x\log x}{e^{x}}= \frac{ (1 + log \: x -x \: log \: x)}{ {e}^{x} }$

Hope it helps you.