# How do you differentiate sin^2(x/6)?

Aug 19, 2015

${y}^{'} = \frac{1}{3} \cdot \sin \left(\frac{x}{6}\right) \cdot \cos \left(\frac{x}{6}\right)$

#### Explanation:

You can differentiate this function

$y = \sin \left(\frac{x}{6}\right)$

by using the chain rule twice, once for ${u}_{1}^{2}$, with ${u}_{1} = \sin \left(\frac{x}{6}\right)$, and once for $\sin {u}_{2}$, with ${u}_{2} = \frac{x}{6}$.

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{{\mathrm{du}}_{1}} \left({u}_{1}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left({u}_{1}\right)$, with

$\frac{d}{\mathrm{dx}} {u}_{1} = \frac{d}{{\mathrm{du}}_{2}} \cdot \sin {u}_{2} \cdot \frac{d}{\mathrm{dx}} \left({u}_{2}\right)$

This will give you

$\frac{d}{\mathrm{dx}} \left({u}_{1}\right) = \cos {u}_{2} \cdot \frac{d}{\mathrm{dx}} \left(\frac{x}{6}\right)$

$\frac{d}{\mathrm{dx}} \left(\sin \left(\frac{x}{6}\right)\right) = \cos \left(\frac{x}{6}\right) \cdot \frac{1}{6}$

Plug this back into your target derivative to get

${y}^{'} = 2 {u}_{1} \cdot \frac{1}{6} \cdot \cos \left(\frac{x}{6}\right)$

${y}^{'} = \textcolor{g r e e n}{\frac{1}{3} \cdot \sin \left(\frac{x}{6}\right) \cdot \cos \left(\frac{x}{6}\right)}$