What is the derivative of sec(x^2)?

Dec 18, 2014

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

Explanation

To solve this question, you would need to use the chain rule (and later on, the quotient rule as well).

In $\sec \left({x}^{2}\right)$, you can quite easily identitfy the "inner" and "outer" functions present to use the chain rule. The "inner" function is ${x}^{2}$, because it is composed/inside of in the $\sec$.

The chain rule is:
$F ' \left(x\right) = f ' \left(g \left(x\right)\right) \left(g ' \left(x\right)\right)$

Or, in words:

the derivative of the outer function (with the inside function left alone!) times the derivative of the inner function.

To make things simpler, I'm going to denote $u = {x}^{2}$ for now because at one point, we will have to leave the inner function alone, but you don't have to do this. If you do, make sure you sub $u = {x}^{2}$ back in at the end so that all your variables are in terms of $x$.

Steps
1) The derivative of the outer function (with the inside function $u$ left alone!)
$\frac{d}{\mathrm{dx}} \sec \left(u\right) = \frac{d}{\mathrm{dx}} \frac{1}{\cos} \left(u\right)$
NB: Unless you have the derivative of $\sec$ memorized, you will have to do a quotient rule here to find its derivative. We know that $\sec \left(u\right) = \frac{1}{\cos} \left(u\right)$ from Trig. So...
d/dx sec (u) = d/dx 1/cos(u) = (cos (u)(0) - 1 (-sin (u)))/( cos (u)cos (u))= sin (u)/ (cos (u) cos (u)) = sin(u)/cos(u) *(1/cos( u)) = tan (u) sec(u)=sec (u) tan (u)

2) The derivative of the inner function:
$\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$

3) Combining the two steps to give the actual derivative:

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