# How do you differentiate given tan^2(x)?

Nov 23, 2016

$\frac{d}{\mathrm{dx}} \left[{\tan}^{2} x\right] = 2 \tan x {\sec}^{2} x$.

#### Explanation:

The big idea here is that we are squaring something. The derivative of ${x}^{2}$ is $2 x$, so the derivative of $u$ squared ($u$ stands for any function) is $2 u$ * (the derivative of $u$).

This is by the chain rule.

In other words, $\frac{d}{\mathrm{dx}} \left[{u}^{2}\right] = 2 u \cdot \frac{d}{\mathrm{dx}} \left[u\right]$.

So $\frac{d}{\mathrm{dx}} \left[{\tan}^{2} x\right] = 2 \tan x \cdot \frac{d}{\mathrm{dx}} \left[\tan x\right]$.

Remember that $\frac{d}{\mathrm{dx}} \left[\tan x\right] = {\sec}^{2} x$, so our final answer is $\frac{d}{\mathrm{dx}} \left[{\tan}^{2} x\right] = 2 \tan x {\sec}^{2} x$.

If you don't remember why or how $\frac{d}{\mathrm{dx}} \left[\tan x\right] = {\sec}^{2} x$, rewrite $\tan x = \sin \frac{x}{\cos} x$ and take the derivative using the quotient rule.

$\frac{d}{\mathrm{dx}} \left[\tan x\right] = \frac{d}{\mathrm{dx}} \left[\sin \frac{x}{\cos} x\right] = \frac{{\cos}^{2} x - \left(- {\sin}^{2} x\right)}{\cos} ^ 2 x = \frac{1}{\cos} ^ 2 x = {\sec}^{2} x$