Question

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

Answer 1

`y'=2x*sec(x^2)tan(x^2)`

Solution

let's `y=f(g(x))`

Using Chain Rule, we get

`y'=f'(g(x))*g'(x)`

for given problem, which is

`y=sec(x^2)`

differentiating with respect to `x` using Chain Rule,

`y'=sec(x^2)tan(x^2)*(x^2)'`

`y'=sec(x^2)tan(x^2)*2x`

`y'=2x*sec(x^2)tan(x^2)`