Derivative f’ of the function f(x)=cos x is: f’(x) = - sin x for any value of x.

## Derivative of cos x

Derivative $f’$ of the function $f(x)=\cos x$ is: $$\forall x \in ]-\infty, +\infty[ , f'(x) = -\sin x$$

## Proof

\begin{aligned} &\frac{\cos (x+h)-\cos x}{h}=\frac{\cos (x) \cos (h)-\sin (x) \sin (h)-\cos x}{h}\\ &\frac{\cos (x+h)-\cos x}{h}=\frac{\sin h}{h} \times(-\sin x)+\cos x \times \frac{\cos h-1}{h}\\ \end{aligned}

We have:

\begin{aligned} \frac{\cos h-1}{h} &=\frac{(\cos h-1)(\cos h+1)}{h(\cos h+1)} \\ &=\frac{\cos ^{2} h-1}{h(\cos h+1)} \\ &=\frac{-\sin ^{2} h}{h(\cos h+1)} \\ &=\frac{\sin h}{h} \times \frac{-\sin h}{\cos h+1} \\ \end{aligned}

Then

$\lim _{h \rightarrow 0} \frac{\cos h-1}{h}=0$

because

$\lim _{h \rightarrow 0} \frac{\sin h}{h}=1$

This equality has been proved in /mathematics/limits/article/proof-of-limit-of-sin-x-x-1-as-x-approaches-0

Now

\begin{aligned} \lim _{h \rightarrow 0}\frac{\cos (x+h)-\cos x}{h}&=\lim _{h \rightarrow 0}\frac{\sin h}{h} \times(-\sin x)+\cos x \times \lim _{h \rightarrow 0}\frac{\cos h-1}{h}\\ &=1\times(-\sin x) + \cos x \times 0 \\ \end{aligned}

We conclude:

$\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h}=-\sin x$