# What is the derivative of y=tan(x) ?

Aug 18, 2014

The derivative of $\tan x$ is ${\sec}^{2} x$.

To see why, you'll need to know a few results. First, you need to know that the derivative of $\sin x$ is $\cos x$. Here's a proof of that result from first principles:

Once you know this, it also implies that the derivative of $\cos x$ is $- \sin x$ (which you'll also need later). You need to know one more thing, which is the Quotient Rule for differentiation:

Once all those pieces are in place, the differentiation goes as follows:

$\frac{d}{\mathrm{dx}} \tan x$
$= \frac{d}{\mathrm{dx}} \sin \frac{x}{\cos} x$

$= \frac{\cos x . \cos x - \sin x . \left(- \sin x\right)}{{\cos}^{2} x}$ (using Quotient Rule)

$= \frac{{\cos}^{2} x + {\sin}^{2} x}{{\cos}^{2} x}$

$= \frac{1}{{\cos}^{2} x}$ (using the Pythagorean Identity)

$= {\sec}^{2} x$