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

## Derivative of sin x

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

## Proof/Demonstration

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

$$\lim _{h \rightarrow 0} \dfrac{\sin h}{h}=1$$ This equality has been proved in /limits/article/proof-of-limit-of-sin-x-x-1-as-x-approaches-0

Now

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

We conclude:

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