Question

$\frac{d}{dx}\left[\frac{\sin x+\cos x}{\sqrt{1+\sin 2x}}\right],\left(0<x<\frac{\pi }{4}\right),=$

• A. 1
• B. 0
• C. -1
• D. 2

Answer 1

Answer: (B)

0

Consider the given function,

$\frac{d}{dx}\left[\frac{\sin x+\mathrm{cosx}}{\sqrt{1+\sin 2x}}\right]$

$=\frac{d}{dx}\left[\frac{\sin x+\mathrm{cosx}}{\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}}\right]$ Since $({\sin }^2\theta +{\cos }^2\theta =1)$ & $\sin 2\theta =2\sin \theta \cos \theta$

$=\frac{d}{dx}\left[\frac{\sin x+\mathrm{cosx}}{\sqrt{{(\sin x+\cos x)}^2}}\right]$

$=\frac{d}{dx}\left[\frac{\sin x+\mathrm{cosx}}{\sin x+\cos x}\right]$

$=\frac{d}{dx}1$

$=0.$

Hence, the value is $0$.