Find, from the first principle, the derivative of $\sqrt {\sin (2x)}$ Find, from the first principle, the derivative of:
$$\sqrt {\sin (2x)}$$
My Attempt:
$$f(x)=\sqrt {\sin (2x)}$$
$$f(x+\Delta x)=\sqrt {\sin (2x+2\Delta x)}$$
Now,
$$f'(x)=\lim_{\Delta x\to 0} \dfrac {f(x+\Delta x)-f(x)}{\Delta x}$$
$$=\lim_{\Delta x\to 0} \dfrac {\sqrt {\sin (2x+2\Delta x)} - \sqrt {\sin (2x)}}{\Delta x}$$
 A: Hint:
Use
$$\dfrac {\sqrt {\sin (2x+2\Delta x)} - \sqrt {\sin (2x)}}{\Delta x}=\dfrac{\sin (2x+2\Delta x)-\sin (2x)}{\sqrt {\sin (2x+2\Delta x)} + \sqrt {\sin (2x)}}\times\dfrac{1}{\Delta x}$$
and $$\sin(a)-\sin(b)=2\sin \dfrac{a-b}{2}\cos\dfrac{a+b}{2}$$
A: Let $f(x)=\sqrt {\sin (2x)}$.
The end goal is to show
$$f'(x)=\frac{\cos(2x)}{\sqrt{\sin(2x)}}$$

\begin{align}
f'(x)
&=\lim_{\delta x\to0}\frac{f(x+\delta x)-f(x)}{\delta x}\\
&=\lim_{\delta x\to0}\frac{\sqrt {\sin (2(x+\delta x))}-\sqrt {\sin (2x)}}{\delta x}\\
&=\lim_{\delta x\to0}\frac{\sqrt {\sin (2(x+\delta x))}-\sqrt {\sin (2x)}}{\delta x}\times\frac{\sqrt {\sin (2(x+\delta x))}+\sqrt {\sin (2x)}}{\sqrt {\sin (2(x+\delta x))}+\sqrt {\sin (2x)}}\\
&=\lim_{\delta x\to0}\frac{1}{\delta x}\frac{\sin (2x+2\delta x)-\sin (2x)}{\sqrt{\sin (2(x+\delta x))}+\sqrt {\sin (2x)}}\\
&=\lim_{\delta x\to0}\frac{1}{\delta x}\frac{2\cos(2x+\delta x)\sin(\delta x)}{\sqrt{\sin (2(x+\delta x))}+\sqrt {\sin (2x)}}\\
&=\lim_{\delta x\to0}\frac{\sin(\delta x)}{\delta x}\cdot\frac{2\lim_{\delta x\to0}\cos(2x+\delta x)}{\lim_{\delta x\to0}\sqrt{\sin (2(x+\delta x))}+\sqrt {\sin (2x)}}\\
&=\lim_{\delta x\to0}\frac{\cos(\delta x)}{1}\cdot\frac{2\cos(2x)}{\sqrt{\sin (2x)}+\sqrt {\sin (2x)}}\\
&=1\cdot\frac{2\cos(2x)}{2\sqrt{\sin (2x)}}\\
&=\frac{\cos(2x)}{\sqrt{\sin (2x)}}\\
\end{align}
