How do you differentiate f'(x) = cos( x² ln(x) ) * ( 2x ln(x) + x ) ?

Oct 10, 2015

"Differentiate" means "find the derivative of". To find $f ' ' \left(x\right)$ use the product and chain rules.

Explanation:

f'(x) = cos( x² ln(x) ) * ( 2x ln(x) + x )

Let's break this into pieces.

$f ' \left(x\right) = u \cdot v$

Where
$u = \cos \left({x}^{2} \ln x\right)$, so

$u ' = - \sin \left({x}^{2} \ln x\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} \ln x\right) = - \sin \left({x}^{2} \ln x\right) \left(2 x \ln x + x\right)$

$v = 2 x \ln x + x$, so

$v ' = 2 \ln x + 2 + 1 = 2 \ln x + 3$

$f ' ' \left(x\right) = u ' v + u v '$

$= {\underbrace{- \sin \left({x}^{2} \ln x\right) \left(2 x \ln x + x\right)}}_{u '} {\underbrace{\left(2 x \ln x + x\right)}}_{v} + {\underbrace{\cos \left({x}^{2} \ln x\right)}}_{u} {\underbrace{\left(2 \ln x + 3\right)}}_{v '}$

$= - \sin \left({x}^{2} \ln x\right) {\left(2 x \ln x + x\right)}^{2} + \left(2 x \ln x + x\right) \left(2 \ln x + 3\right)$

This can be rewritten in several ways using algebra.

Note
The much simpler question is to find the antiderivative, that is, to integrate f'(x) = cos( x² ln(x) ) * ( 2x ln(x) + x )

Because $\left(2 x \ln \left(x\right) + x\right)$ is thederivative of the argument of $\cos$, we see that this is $f ' \left(x\right) = \cos u \frac{\mathrm{du}}{\mathrm{dx}}$

So, $f \left(x\right) = \sin u + C = \sin \left({x}^{2} \ln x\right) + C$