Answer
See proof
Work Step by Step
First we have to determine $\dfrac{d}{dx} (\cos x)$.
We have:
$\dfrac{d}{dx} (\cos x)=\lim\limits_{h \to 0}\dfrac{\cos (x+h)-\cos x}{h}$
Use the identity:
$\cos(x+y)=\cos x\cos y-\sin x\sin y$
$\dfrac{d}{dx} (\cos x)=\lim\limits_{h \to 0}\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h}$
$=\lim\limits_{h \to 0}\dfrac{\cos x(\cos h-1)-\sin x\sin h}{h}$
$=\lim\limits_{h \to 0}\dfrac{\cos x(\cos h-1)}{h}-\lim\limits_{h \to 0}\dfrac{\sin x\sin h}{h}$
$=\cos x\lim\limits_{h \to 0}\dfrac{\cos h-1}{h}-\sin x\lim\limits_{h \to 0}\dfrac{\sin h}{h}$
We use the limits:
$\lim\limits_{x \to 0}\dfrac{\cos x-1}{x}=0$
$\lim\limits_{x \to 0}\dfrac{\sin x}{x}=1$
$\dfrac{d}{dx} (\cos x)=\cos x\cdot 0-\sin x\cdot 1=\sin x$
We got:
$\dfrac{d}{dx} (\cos x)=-\sin x$
