# Derivative of tan(x)

## Statement

The derivative of $\tan(x)$ is:

$$\tfrac{d}{dx}\bigg(\tan(x)\bigg) = 1 + \tan(x)^2$$

## Proof

Define function $f$.

$$f(x) = \tan(x)$$

Rewrite $f$ by using tangent is sine divided by cosine.

$$f(x) = \frac{\sin(x)}{\cos(x)}$$

Now differentiate this function with the quotient rule, $\frac{d}{dx}\big(\sin(x)\big) = \cos(x)$ and $\frac{d}{dx}\big(\cos(x)\big) = -\sin(x)$.

$$f'(x) = \frac{\cos(x)* \cos(x) - \sin(x)* - \sin(x)}{\cos(x)^2}$$

Simplify this function and split the fraction.

$$f'(x) = \frac{\cos(x)^2 + \sin(x)^2}{\cos(x)^2} = \frac{\cos(x)^2}{\cos(x)^2} + \frac{\sin(x)^2}{\cos(x^2)}$$

Finally, use that tangent is sine divided by cosine.

$$f'(x) = 1 + \tan(x)^2$$