# What is the derivative of y=sec^2(x) + tan^2(x)?

Aug 23, 2014

The derivative of $y = {\sec}^{2} x + {\tan}^{2} x$ is:

$4 {\sec}^{2} x \tan x$

Process:

Since the derivative of a sum is equal to the sum of the derivatives, we can just derive ${\sec}^{2} x$ and ${\tan}^{2} x$ separately and add them together.

For the derivative of ${\sec}^{2} x$, we must apply the Chain Rule:

$F \left(x\right) = f \left(g \left(x\right)\right)$
$F ' \left(x\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$,

with the outer function being ${x}^{2}$, and the inner function being $\sec x$. Now we find the derivative of the outer function while keeping the inner function the same, then multiply it by the derivative of the inner function. This gives us:

$f \left(x\right) = {x}^{2}$
$f ' \left(x\right) = 2 x$

$g \left(x\right) = \sec x$
$g ' \left(x\right) = \sec x \tan x$

Plugging these into our Chain Rule formula, we have:

$F ' \left(x\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$,
$F ' \left(x\right) = 2 \left(\sec x\right) \sec x \tan x = 2 {\sec}^{2} x \tan x$

Now we follow the same process for the ${\tan}^{2} x$ term, replacing $\sec x$ with $\tan x$, ending up with:

$f \left(x\right) = {x}^{2}$
$f ' \left(x\right) = 2 x$

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

$F ' \left(x\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$,
$F ' \left(x\right) = 2 \left(\tan x\right) {\sec}^{2} x = 2 {\sec}^{2} x \tan x$

$2 {\sec}^{2} x \tan x + 2 {\sec}^{2} x \tan x$
= $4 {\sec}^{2} x \tan x$