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, $$f'(x)=\lim_{\Delta x\to 0} \dfrac {f(x+\Delta x)-f(x)}{\Delta x}$$ $$=\lim_{\Delta x\to 0} \dfrac {\sqrt {\sin (2x+2\Delta x)} - \sqrt {\sin (2x)}}{\Delta x}$$
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$\begingroup$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 $$\sin(a)-\sin(b)=2\sin \dfrac{a-b}{2}\cos\dfrac{a+b}{2}$$
$\endgroup$ 1 $\begingroup$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}
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