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