Question

Find the derivatives of the following functions.
(a) $\sqrt{\ln x}$ (b) $\sqrt{\sin 2x}$ (c) $(\sin 2x+8{)}^3$

Answer 1

(a) $\sqrt{\ln x}$

$\frac{d}{dx}\sqrt{\ln x}=\frac{1}{2\sqrt{\ln x}}\frac{1}{x}$ [Apply Chain Rule : $\frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}]$.

$=\frac{1}{2x\sqrt{\ln x}}$

(b) $\sqrt{\sin 2x}$

$\frac{d}{dx}\sqrt{\sin 2x}=\frac{\cos 2x.\left(2\right)}{2\sqrt{\sin 2x}}$

$=\frac{\cos 2x}{\sqrt{\sin 2x}}=\frac{\cos 2x}{\sin 2x}\sqrt{\sin 2x}$

$=\cot 2x\sqrt{\sin 2x}$

(c) $\sqrt{(\sin 2x+8{)}^3}$

$\frac{d}{dx}\sqrt{(\sin 2x+8{)}^3}=\frac{3(\sin 2x+8{)}^2}{2\sqrt{(\sin 2x+8{)}^3}}\left(2\cos 2x\right)$

$=\frac{3(\sin 2x+8)\left(\cos 2x\right)}{\sqrt{\sin 2x+8}}$

$=3\sqrt{\sin 2x+8}\left(\cos 2x\right)$.