# How do you find the derivative of y = 6 cos(x^3 + 3) using the chain rule?

Nov 11, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 18 {x}^{2} \sin \left({x}^{3} + 3\right)$

#### Explanation:

$\text{Let } u = {x}^{3} + 3 \to \frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2}$

$\text{Let } v = \cos \left(u\right) \to \frac{\mathrm{dv}}{\mathrm{du}} = - \sin \left(u\right)$

$\text{Let } y = 6 v \to \frac{\mathrm{dy}}{\mathrm{dv}} = 6$

Target is $\frac{\mathrm{dy}}{\mathrm{dx}}$

By cancelling out $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \times \frac{\mathrm{dv}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(6\right) \times \left\{- \sin \left(u\right)\right\} \times \left(3 {x}^{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(6\right) \times \left(- 1\right) \times \left(3\right) \times \left\{\sin \left(u\right)\right\} \times \left\{{x}^{2}\right\}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 18 {x}^{2} \sin \left({x}^{3} + 3\right)$