Find derivative of square root sin x by first principle.

$f\text{'}\left(x\right)=\underset{h\to 0}{lim}\frac{\sin }{}\frac{(x+h)}{-\sin x}$

Using $\sin (A+B)=\sin A\cos B+\sin B\cos A$ we get

$f\text{'}\left(x\right)=\underset{h\to 0}{lim}\frac{\sin x\cos h}{}\frac{+\sin h\cos x}{-\sin x}$
$f\text{'}\left(x\right)=\underset{h\to 0}{lim}\frac{\sin x}{}\frac{(\cos h-1)+}{\sin h\cos x}$
$f\text{'}\left(x\right)=\underset{h\to 0}{lim}(\frac{\sin x(\cos h-1)}{h}+\frac{\sin h\cos x}{h})$

$f\text{'}\left(x\right)=\underset{h\to 0}{lim}\frac{\sin x}{}\frac{(\cos h-1)}{h}+\underset{h\to 0}{lim}\frac{\sin h\cos x}{h}$

$f\text{'}\left(x\right)=\left(\sin x\right)\underset{h\to 0}{lim}\frac{\cos h-1}{h}+\left(\cos x\right)\underset{h\to 0}{lim}\frac{\sin h}{h}$

We know have to rely on some standard limits:

$\underset{h\to 0}{lim}\frac{\sin h}{h}=1$
$\underset{h\to 0}{lim}\frac{\cos h-1}{h}=0$

And so using these we have:

$f\text{'}\left(x\right)=0+\left(\cos x\right)\left(1\right)=\cos x$

Hence,

$\frac{d}{dx}\sin x=\cos x$