# What is the derivative of y = sin(tan(3x))?

Aug 11, 2015

${y}^{'} = 3 \cos \left(\tan \left(3 x\right)\right) \cdot {\sec}^{2} \left(3 x\right)$

#### Explanation:

You can differentiate this function by using the chain rule three times.

First, start by writing your function as $y = \sin u$, where $u = \tan \left(3 x\right)$. Its derivative will take the form

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

$\frac{d}{\mathrm{dx}} \left(\sin u\right) = \cos u \cdot \frac{d}{\mathrm{dx}} \left(\tan \left(3 x\right)\right)$

Now focus on $\frac{d}{\mathrm{dx}} \left(\tan \left(3 x\right)\right)$, which can be written as $\tan \left({u}_{1}\right)$, with ${u}_{1} = 3 x$.

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

$\frac{d}{\mathrm{dx}} \left(\tan {u}_{1}\right) = {\sec}^{2} {u}_{1} \cdot \frac{d}{\mathrm{dx}} \left(3 x\right)$

$\frac{d}{\mathrm{dx}} \left(\tan \left(3 x\right)\right) = {\sec}^{2} \left(3 x\right) \cdot 3$

Plug this into your target derivative to get

$\frac{d}{\mathrm{dx}} \left(\sin \left(\tan \left(3 x\right)\right)\right) = \cos \left(\tan \left(3 x\right)\right) \cdot 3 {\sec}^{2} \left(3 x\right)$

Therefore,

${y}^{'} = \frac{d}{\mathrm{dx}} \left(\sin \left(\tan \left(3 x\right)\right)\right) = \textcolor{g r e e n}{3 \cos \left(\tan \left(3 x\right)\right) \cdot {\sec}^{2} \left(3 x\right)}$