# What is the derivatives of sec2x and tan2x?

Mar 15, 2015

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

Use the derivatives of trig functions and the chain rule:

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

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

Explanation
You'll need the derivatives of $y = \sec x$ and $y = \tan x$
$\frac{d}{\mathrm{dx}} \left(\sec x\right) = \sec x \tan x$. and $\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x$

You'll also want the Chain Rule.

There are various notations for derivatives and the Chain Rule, but for this question, this is a good one:
Suppose that we know $\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = f ' \left(x\right)$, then if we want $\frac{d}{\mathrm{dx}} \left(f \left(u\right)\right)$ , the Chain Rule tells us to find the derivative of the outside function (that's $f '$) and evaluate it, not at $x$, but at $u$. Then multiply by the derivative of the inside.

The Chain Rule:
$\frac{d}{\mathrm{dx}} \left(f \left(u\right)\right) = f ' \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Finding $\frac{d}{\mathrm{dx}} \left(\tan \left(2 x\right)\right)$

The outside function is $\tan$ and the inside function (the $u$) is $2 x$.

The derivative of the tangent function is the square of the secant function.
$\frac{d}{\mathrm{dx}} \left(f \left(u\right)\right) = \frac{d}{\mathrm{dx}} \left(\tan \left(u\right)\right) = {\sec}^{2} \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

As you get more experience, you'll simply write:

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

Finding $\frac{d}{\mathrm{dx}} \left(\sec \left(2 x\right)\right)$

The outside function is $\sec$ and the inside function (the $u$) is $2 x$.

The derivative of $f \left(x\right) = \sec x$ is the function (singular), $f ' \left(x\right) = \sec x \tan x$.

So the derivative of $f \left(2 x\right)$ is $f ' \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x\right)$, we write:

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