Question

$\frac{d}{dx}\left[\sqrt{\cos \sqrt{x}}\right]=$

• A. $\frac{\cos \sqrt{x}}{4\sqrt{x}\sqrt{\sin \sqrt{x}}}$
• B. $\frac{-\sin \sqrt{x}}{4\sqrt{x}\sqrt{\cos \sqrt{x}}}$
• C. $\sin \sqrt{x}$
• D. $\cos \sqrt{x}$

Answer 1

The correct option is B $\frac{-\sin \sqrt{x}}{4\sqrt{x}\sqrt{\cos \sqrt{x}}}$

Using chain rule:
$\frac{d}{dx}\left[\sqrt{\cos \sqrt{x}}\right]=\frac{1}{2}[\cos \sqrt{x}{]}^{\frac{1}{2}-1}\times \frac{d}{dx}(\cos \sqrt{x})$
$=\frac{1}{2\sqrt{\cos \sqrt{x}}}.(-\sin \sqrt{x})\times \frac{d}{dx}\sqrt{x}=\frac{-\sin \sqrt{x}}{2\sqrt{\cos \sqrt{x}}}\times \frac{1}{2\sqrt{x}}=\frac{-\sin \sqrt{x}}{4\sqrt{x}\sqrt{\cos \sqrt{x}}}$