How do you differentiate f(x) = sin ( x² ln(x) )?

Jan 29, 2016

$f ' \left(x\right) = \left(2 x \ln \left(x\right) + x\right) \cos \left({x}^{2} \ln \left(x\right)\right)$

Explanation:

To find the derivative of a sine function like this, we will have to use the chain rule:

$\frac{d}{\mathrm{dx}} \left(\sin \left(u\right)\right) = \cos \left(u\right) \cdot u '$

In this instance, $u = {x}^{2} \ln \left(x\right)$, so differentiation yields

$f ' \left(x\right) = \cos \left({x}^{2} \ln \left(x\right)\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} \ln \left(x\right)\right)$

To differentiate ${x}^{2} \ln \left(x\right)$, the product rule is necessary. Recall that the derivative on $\ln \left(x\right)$ is $\frac{1}{x}$.

$\frac{d}{\mathrm{dx}} \left({x}^{2} \ln \left(x\right)\right) = \ln \left(x\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right) + {x}^{2} \frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) = 2 x \ln x + x$

Plugging this back in, we see that

$f ' \left(x\right) = \left(2 x \ln \left(x\right) + x\right) \cos \left({x}^{2} \ln \left(x\right)\right)$