Differentiate x^2 e^x logx

Question

asked Oct 15, 2021 3.9k views

1 Answer

Lyncean Patel · Answer 1 · 2021-10-19T07:11:58+0000

Product rule:

$(\mathrm d)/(\mathrm dx)(x^2e^x\log x)$

$=(\mathrm d(x^2))/(\mathrm dx)e^x\log x+x^2(\mathrm d(e^x))/(\mathrm dx)\log x+x^2e^x(\mathrm d(\log x))/(\mathrm dx)$

Power rule:

$(\mathrm d(x^2))/(\mathrm dx)=2x$

The exponential function is its own derivative:

$(\mathrm d(e^x))/(\mathrm dx)=e^x$

Assuming the base of
$\log x$ is
, its derivative is

$(\mathrm d(\log x))/(\mathrm dx)=\frac1x$

But if you mean a logarithm of arbitrary base
, we have

$y=\log_bx\implies x=b^y=e^(y\ln b)\implies1=e^(y\ln b)\ln b(\mathrm dy)/(\mathrm dx)$

$\implies(\mathrm dy)/(\mathrm dx)=(e^(-y\ln b))/(\ln b)=\frac1{b^y\ln b}$

$\implies(\mathrm d(\log_bx))/(\mathrm dx)=\frac1{x\ln b}$

So we end up with

$2xe^x\log x+x^2e^x\log x+\frac{x^2e^x}x$

$=xe^x(2\log x+x\log x+1)$