# What is the derivative of z=sin^3(theta)?

Aug 10, 2015

$\frac{\mathrm{dz}}{d \theta} = 3 {\sin}^{2} \left(\theta\right) \cos \left(\theta\right)$

#### Explanation:

This follows from the Chain Rule: $\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

For the function ${\sin}^{3} \left(\theta\right)$, if we let $g \left(\theta\right) = \sin \left(\theta\right)$ and $f \left(\theta\right) = {\theta}^{3}$, then ${\sin}^{3} \left(\theta\right) = f \left(g \left(\theta\right)\right)$.

Since $f ' \left(\theta\right) = 3 {\theta}^{2}$ and $g ' \left(\theta\right) = \cos \left(\theta\right)$, we get:

$\frac{\mathrm{dz}}{d \theta} = f ' \left(g \left(\theta\right)\right) \cdot g ' \left(\theta\right) = 3 {\sin}^{2} \left(\theta\right) \cdot \cos \left(\theta\right)$.