Question

$\frac{x lnx}{e^x}$ Differentiate the following functions w.r.t. x

Answer 1

Solution:

To differentiate $\frac{xln(x)}{e^{x} }$ with respect to x.

We can rewrite $\frac{xln(x)}{e^{x} }$ as $xe^{-x}ln(x)$

$\frac{d}{dx} [xe^{-x}ln(x)]$

[Apply Product Rule => [uvw]' = u'vw + uv'w + uvw']

$\frac{d}{dx} [x] . e^{-x} ln(x) + x.\frac{d}{dx}[e^{-x}].ln(x) + xe^{-x} . \frac{d}{dx}[ln(x)]$

$1e^{-x} ln(x) + xe^{-x}.\frac{d}{dx}[-x].ln(x) + xe^{-x} .\frac{1}{x}$

$xe^{-x} (-\frac{d}{dx}[x])ln(x) + e^{-x}ln(x) + e^{-x}$

$-xe^{-x}. 1ln(x) + e^{-x} ln(x) + e^{-x}$

$-xe^{-x} ln(x) + e^{-x}ln(x) + e^{-x}$

We can rewrite the answer as

$-e^{-x} ((x-1)ln(x)-1)$

The differentiation of $\frac{xln(x)}{e^{x} }$ is $-xe^{-x} ln(x) + e^{-x}ln(x) + e^{-x}$ or $-e^{-x} ((x-1)ln(x)-1)$.

Extra Information:

The sign [ ' ] means single time differentiation.

As there are more signs number of differentiation times increases.