Question

Differentiate from first principle:

$\left(i\right)\sqrt{\sin 2x}$

Answer 1

Given:

$f\left(x\right)=\sqrt{\sin 2x}$

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Putting $f\left(x\right)$ in above expression, we get:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sqrt{\sin (2x+2h)}-\sqrt{\sin 2x}}{h}$

Rationalizing the numerator, we get:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sqrt{\sin (2x+2h)}-\sqrt{\sin 2x}}{h}\times \frac{\sqrt{\sin (2x+2h)}+\sqrt{\sin 2x}}{\sqrt{\sin (2x+2h)}+\sqrt{\sin 2x}}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sin (2x+2h)-\sin 2x}{h(\sqrt{\sin (2x+2h)}+\sqrt{\sin 2x})}$

Applying the formula,

$\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right),$ we get

$\Rightarrow f^{\prime }\left(x\right)$

$=\underset{h\to 0}{lim}\frac{2\cos \left(\frac{2x+2h+2x}{2}\right)\sin \left(\frac{2x+2h-2x}{2}\right)}{h(\sqrt{\sin (2x+2h)}+\sqrt{\sin 2x})}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{2\cos (2x+h)}{(\sqrt{\sin (2x+2h)}+\sqrt{\sin 2x})}.\frac{\sin h}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{2\cos (2x+0)}{(\sqrt{\sin (2x+0)}+\sqrt{\sin 2x})}$

$[\because \underset{h\to 0}{lim}\frac{\sin h}{h}=1]$

$\Rightarrow f^{\prime }\left(x\right)=\frac{\cos 2x}{\sqrt{\sin 2x}}$