$(\sin x)'=\cos x$

$f(x)=\sin x$;

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{\sin(x+h)-\sin x}{h}$

We use the formula for the sine of the sum of two angles [??]

$\sin(x+h)-\sin x=\sin x\cos h + \cos x\sin h-\sin x=\sin x(\cos h -1)+ \cos x\sin h$

And now

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{\sin(x+h)-\sin x}{h}=\sin x \lim_{h\longrightarrow 0}\dfrac{\cos h-1}{h}+\cos x \lim_{h\longrightarrow 0}\dfrac{\sin h}{h}=\sin x \times 0 + \cos x \times 1=\cos x$

having used the derivatives of the trigonometric functions in the origin

$(\cos x)'=-\sin x$

$f(x)=\cos x$;

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{\cos(x+h)-\cos x}{h}$

We use the formula for the cosine of the sum of two angles [??]

$\cos(x+h)-\cos x=\cos x\cos h - \sin x\sin h-\cos x=\cos x(\cos h -1)- \sin x\sin h$

And now

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{\cos(x+h)-\cos x}{h}=\cos x \lim_{h\longrightarrow 0}\dfrac{\cos h-1}{h}-\sin x \lim_{h\longrightarrow 0}\dfrac{\sin h}{h}=\cos x \times 0 - \sin x \times 1=-\sin x$

having used again the derivatives of the trigonometric functions in the origin