# What is the second derivative of f(x)=tan(3x)?

Dec 14, 2015

$f ' ' \left(x\right) = 18 {\sec}^{2} \left(3 x\right) \tan \left(3 x\right)$

#### Explanation:

We will use the chain rule, together with the derivatives:

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

• $\frac{d}{\mathrm{dx}} 3 x = 3$

• $\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$

• $\frac{d}{\mathrm{dx}} \sec \left(x\right) = \sec \left(x\right) \tan \left(x\right)$

First Derivative:
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \tan \left(3 x\right)$

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

$= 3 {\sec}^{2} \left(3 x\right)$

Second Derivative:

$f ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(f ' \left(x\right)\right)$

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

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

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

$= 6 \sec \left(3 x\right) \left(\sec \left(3 x\right) \tan \left(3 x\right)\right) \left(\frac{d}{\mathrm{dx}} 3 x\right)$

$= 18 {\sec}^{2} \left(3 x\right) \tan \left(3 x\right)$