# Derivative of tan x

The first derivative of $\tan (x)$ is calculated using the derivative of $\sin x$ and $\cos x$ and the quotient rule of derivatives.

## Calculation of the Derivative of tan x

A trigonometric identity relating $\tan x$, $\sin x$ and $\cos x$ is given by $\tan x = \dfrac { \sin x }{ \cos x }$ One way to find the derivative of $\tan x$ is to use the quotient rule of differentiation; hence
$\displaystyle {\dfrac {d}{dx} \tan x = \dfrac {d}{dx} (\dfrac{\sin x }{\cos x}) = (\dfrac{{ (\dfrac {d}{dx}\sin x) }{ \cos x } - \sin x (\dfrac {d}{dx} \cos x) }{\cos^2 x}}$

Use the formulae for the
derivative of the trigonometric functions $\sin x$ and $\cos x$ given by $\dfrac {d}{dx}\sin x = \cos x$ and $\dfrac {d}{dx}\cos x = - \sin x$ and substitute to obtain

$\displaystyle {\dfrac {d}{dx} \tan x = (\dfrac{{ \cos x \cos x } - \sin x (-\sin x) }{\cos^2 x}}$

Simplify

$\displaystyle {= \dfrac{ \cos^2 x + \sin^2 x } {\cos^2 x} = \dfrac{ 1 }{\cos^2 x} = \sec^2 x }$
conclusion
$\displaystyle {\dfrac {d}{dx} \tan x = \sec^2 x}$

## Graph of tan x and its Derivative

The graphs of $\tan(x)$ and its derivative are shown below.

## Derivative of the Composite Function tan (u(x))

We now have a composite function which is a function (tan) of another function (u). Use the chain rule of differentiation to write

$\displaystyle \dfrac{d}{dx} \tan (u(x)) = (\dfrac{d}{du} \tan u) (\dfrac{d}{dx} u )$

Simplify

$= \sec^2 u \dfrac{d}{dx} u$

Conclusion

$\displaystyle \dfrac{d}{dx} \tan (u(x)) = \sec^2 u \dfrac{d}{dx} u$

Example 1
Find the derivative of the composite tan functions

1. $f(x) = \tan (2x -1)$
2. $g(x) = \tan (\cos(x))$
3. $h(x) = \tan (\dfrac{x-1}{x+2})$

Solution to Example 1

1. Let $u(x) = 2x - 1$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} (2x - 1) = 2$ and apply the rule

$\displaystyle \dfrac{d}{dx} f(x) = (\sec^2 u \dfrac{d}{dx} u = \sec^2 (2x-1) \times 2 = 2 \sec^2 (2x-1) )$

2. Let $u(x) = \cos x$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} \cos x = - \sin x$ and apply the rule

$\displaystyle \dfrac{d}{dx} g(x) = (\sec^2 u \dfrac{d}{dx} u = \sec^2 (\cos x) \times (- \sin x) = - \sin x \sec^2 (\cos x)$

3. Let $u(x) = \dfrac{x-1}{x+2}$ and therefore $\dfrac{d}{dx} u = \dfrac{3}{(x+2)^2}$ and apply the rule obtained above

$\displaystyle \dfrac{d}{dx} h(x) = \sec^2 u \dfrac{d}{dx} u = \sec^2 (\dfrac{x-1}{x+2}) \times \dfrac{3}{(x+2)^2} = \dfrac {3 \sec^2 (\dfrac{x-1}{x+2})}{(x+2)^2}$