# What is the Derivative of y=sec(x^2)?

Aug 13, 2014

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

Solution

let's $y = f \left(g \left(x\right)\right)$

Using Chain Rule, we get

$y ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

for given problem, which is

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

differentiating with respect to $x$ using Chain Rule,

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

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

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