# How do you find the derivative of sin(x^3)?

May 29, 2015

We use the chain rule.

http://socratic.org/calculus/basic-differentiation-rules/chain-rule

Using the notation provided there, if we define $y \left(x\right) = \sin \left({x}^{3}\right)$ and $u = {x}^{3}$, we may rewrite $y \left(x\right)$ as $y \left(u\right) = \sin \left(u\right)$

From the chain rule we know that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \frac{\mathrm{dy}}{\mathrm{du}}$. Recall that $u \left(x\right) = {x}^{3}$ and $y \left(u\right) = \sin \left(u\right)$. Therefore, by the power rule, $\frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2}$, and by the definitions of trigonometric derivatives, $\frac{\mathrm{dy}}{\mathrm{du}} = \cos \left(u\right)$. Thus:

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

Substituting ${x}^{3}$ back for u yields:

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