# How do you differentiate  y =cos^3(5x^2-2) using the chain rule?

Jun 15, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 30 x \setminus {\cos}^{2} \left(5 {x}^{2} - 2\right) \setminus \sin \left(5 {x}^{2} - 2\right)$

#### Explanation:

We have:

$y = {\cos}^{3} \left(5 {x}^{2} - 2\right)$

We will need two application of the chain rule:

$\frac{d}{\mathrm{dx}} {u}^{3} \setminus \setminus \setminus = 3 {u}^{2} \frac{\mathrm{du}}{\mathrm{dx}}$
$\frac{d}{\mathrm{dx}} \cos v = - \sin v \frac{\mathrm{dv}}{\mathrm{dx}}$

Thus we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\cos}^{2} \left(5 {x}^{2} - 2\right) \cdot \left(- \sin \left(5 {x}^{2} - 2\right)\right) \cdot \left(10 x\right)$
$\setminus \setminus \setminus \setminus \setminus = - 30 x \setminus {\cos}^{2} \left(5 {x}^{2} - 2\right) \setminus \sin \left(5 {x}^{2} - 2\right)$