# How do you find the derivative of y=tan(x) ?

Aug 1, 2014

$y = \tan x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} x$

This is a common identity which many people memorize, along with the derivatives of $\sin x$ and $\cos x$ because they appear so frequently.

It can be proven easily using two well-known trig identities and the quotient rule.

Process:
First recall from pre-calculus that $\tan \theta = \sin \frac{\theta}{\cos} \theta$.

So, we can rewrite $y = \tan x$ equivalently as:

$y = \sin \frac{x}{\cos} x$

Now, we differentiate, and apply the quotient rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{d}{\mathrm{dx}} \left[\sin x\right] \cdot \cos x - \frac{d}{\mathrm{dx}} \left[\cos x\right] \cdot \sin x}{\cos x} ^ 2$

We know that the derivative of $\sin x$ is $\cos x$, and that the derivative of $\cos x$ is $- \sin x$. So, upon simplifying the above equation, we arrive at:

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

It should be clear that the numerator can be simplified using another trig identity.

Recall ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$. (or the Pythagorean identity)

Substitution yields:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\cos}^{2} x}$

which is equivalent to:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} x$