# What is the second derivative of f(x)= secx^2?

Feb 26, 2018

$f ' ' \left(x\right) = 8 {x}^{3} {\sec}^{3} {x}^{2} \tan {x}^{2}$

#### Explanation:

Original Function:
$f \left(x\right) = \sec {x}^{2}$

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

Remember to use chain rule for the ${x}^{2}$ inside the argument of the trig function!

First derivative:
$f ' \left(x\right) = 2 x \sec {x}^{2} \tan {x}^{2}$

Second derivative:
$f ' ' \left(x\right) = 2 x \left(2 x\right) \sec {x}^{2} \tan {x}^{2} \left(2 x\right) {\sec}^{2} {x}^{2}$
$f ' ' \left(x\right) = 8 {x}^{3} \sec {x}^{2} \tan {x}^{2} {\sec}^{2} {x}^{2}$
$f ' ' \left(x\right) = 8 {x}^{3} {\sec}^{3} {x}^{2} \tan {x}^{2}$