# What is the derivative of  sin^2(x) cos (x)?

Nov 21, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin x \left(3 {\cos}^{2} x - 1\right)$

#### Explanation:

$y = \left(1 - {\cos}^{2} x\right) \cos x = \cos x - {\cos}^{3} x$

We know the derivative of $\cos x$ is $- \sin x$. Letting $y = {u}^{3}$ and $u = \cos x$, we have: $\left({\cos}^{3} x\right) ' = - \sin x 3 {u}^{2} = - \sin x 3 {\left(\cos x\right)}^{2} = - 3 {\cos}^{2} x \sin x$

The derivative of the entire expression is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sin x - \left(- 3 {\cos}^{2} x \sin x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\cos}^{2} x \sin x - \sin x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin x \left(3 {\cos}^{2} x - 1\right)$

Hopefully this helps!

Nov 21, 2016

$\sin x \left(3 {\cos}^{2} x - 1\right)$

#### Explanation:

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

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

= $- {\sin}^{3} x + 2 \sin x {\cos}^{2} x$

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

=$\sin x \left({\cos}^{2} x - 1 + 2 {\cos}^{2} x\right)$

= $\sin x \left(3 {\cos}^{2} x - 1\right)$