# How do you find the derivative of h(x) = cos(4x^3)?

Aug 14, 2015

${h}^{'} = - 12 {x}^{2} \cdot \sin \left(4 {x}^{3}\right)$

#### Explanation:

You can differentiate this function by using the chain rule for $\cos u$, with $u = 4 {x}^{3}$.

You need to know that

$\frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x$

So, the derivative of $h \left(x\right)$ will be

$\frac{d}{\mathrm{dx}} \left(\cos u\right) = \left[\frac{d}{\mathrm{du}} \cos u\right] \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left(\cos u\right) = - \sin u \cdot \frac{d}{\mathrm{dx}} \left(4 {x}^{3}\right)$

$\frac{d}{\mathrm{dx}} \left(\cos \left(4 {x}^{3}\right)\right) = - \sin \left(4 {x}^{3}\right) \cdot 12 {x}^{2}$

$\frac{d}{\mathrm{dx}} \left(\cos \left(4 {x}^{3}\right)\right) = \textcolor{g r e e n}{- 12 {x}^{2} \cdot \sin \left(4 {x}^{3}\right)}$