# How do you find the derivative of y = cos^3(3x+1)?

Oct 25, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 9 \sin \left(3 x + 1\right) {\cos}^{2} \left(3 x + 1\right)$

#### Explanation:

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

Applying chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = \cos \left(3 x + 1\right) ,$ then $\frac{\mathrm{du}}{\mathrm{dx}} = - 3 \sin \left(3 x + 1\right)$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{d}{\mathrm{du}} {u}^{3} = 3 {u}^{2} = 3 {\cos}^{2} \left(3 x + 1\right)$

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

$= - 9 \sin \left(3 x + 1\right) {\cos}^{2} \left(3 x + 1\right)$