# Proof of Derivative of cos x

The definition of the derivative is used to prove the formula for the derivative of $\cos (x)$ . The derivative of a cosine composite function is also presented including examples with their solutions.

## Proof of the Derivative of cos x Using the Definition

The definition of the derivative $f'$ of a function $f$ is given by
$f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$
Let $f(x) = \cos(x)$ and write the derivative of $\cos(x)$ as a limit
$f'(x) = \lim_{h \to 0} \dfrac{\cos(x+h)-\cos(x)}{h}$
Use the formula $\cos(x+h) = \cos(x)\cos(h) - \sin(x)\sin(h)$ to rewrite the derivative of $cos(x)$ as
$f'(x) = \lim_{h \to 0} \dfrac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h}$
Rewrite $f'(x)$ as follows
$f'(x) = \lim_{h \to 0} \dfrac{\cos(x) (cos(h) - 1) - \sin(x) \sin(h))}{h}$
Use the theorem on limits that states: the limit of a difference of two functions is equal to the difference of the limits, to rewrite $f'(x)$ as follows
$f'(x) = \lim_{h \to 0} \dfrac{\cos(x) (cos(h) - 1)}{h} - \lim_{h \to 0} \dfrac{\sin(x) \sin(h)}{h}$
Rewrite the above as
$f'(x) = \cos(x) \lim_{h \to 0} \dfrac{ (cos(h) - 1)}{h} - \sin(x) \lim_{h \to 0} \dfrac{ \sin(h)}{h}$

We now use the well known results on the limits of trigonometric functions
$\lim_{h \to 0} \dfrac{\sin(h)}{h} = 1$ , prooved in the
use of squeezing theorem to find limits of mathematical functions .
$\lim_{h \to 0} \dfrac{cos(h) - 1}{h} = 0$ , prooved in
calculate limits of trigonometric functions
to simplify $f'(x)$ to
$f'(x) = \sin(x) (0) - \sin(x) (1) = - sin(x)$
conclusion
$\displaystyle {\dfrac {d}{dx} \cos x = - \sin x }$

## Graph of cos x and its Derivative

The graphs of $\cos(x)$ and its derivative are shown below. Note that at any maximum or minimum of $\cos(x)$ corresponds a zero of the derivative. For any interval over which $\cos(x)$ is increasing the derivative is positive and for any interval over which $\cos(x)$ is decreasing, the derivative is negative.

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

We now consider the composite function cos of another function u(x). Use the chain rule of differentiation to write

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

Simplify

$= - \sin u \dfrac{d}{dx} u$

Conclusion

$\displaystyle \dfrac{d}{dx} \cos (u(x)) = - \sin u \dfrac{d}{dx} u$

Example 1
Find the derivative of the composite cos functions

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

Solution to Example 1

1. Let $u(x) = 2x+2$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} (2x+2) = 2$ and apply the rule for the composite cos function given above

$\displaystyle \dfrac{d}{dx} f(x) = \dfrac{d}{dx} \cos(u) = -\sin u \dfrac{d}{dx} u = -\sin (2x+2) \times 2$

$= - 2 \sin (2x+2)$

2. Let $u(x) = \tan x$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} \tan x = \sec^2 x$ and apply the above rule of differentiation for the composite cos function

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

$= - \sec^2 x \; \sin (\tan x)$

3. Let $u(x) = \left(\dfrac{x^2}{x^2+1} \right)$ and therefore $\dfrac{d}{dx} u = \frac{2x}{\left(x^2+1\right)^2}$ and apply the rule of differentiation for the composite cos function obtained above

$\displaystyle \dfrac{d}{dx} h(x) = \dfrac{d}{dx} \cos(u) = - \sin u \dfrac{d}{dx} u = - \sin (\frac{2x}{\left(x^2+1\right)^2}) \times ( \dfrac{2x}{\left(x^2+1\right)^2})$

$= - \dfrac{2x}{\left(x^2+1\right)^2} \cos \left(\dfrac{x^2}{x^2+1} \right)$