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Method of Differentiation: Question

JEE Maths question with a full step-by-step solution.

Question
ddx[sin⁡x+cos⁡x1+sin⁡2x]\dfrac{d}{dx}\left[\dfrac{\sin x + \cos x}{\sqrt{1 + \sin 2x}}\right], where 0<x<π40 < x < \dfrac{\pi}{4}, is:
A11
B00correct
C−1-1
D22
Solution
Step 1: Simplify the denominator Since 1+sin⁡2x=(sin⁡x+cos⁡x)21 + \sin 2x = (\sin x + \cos x)^2, and sin⁡x+cos⁡x>0\sin x + \cos x > 0 for x∈(0,π4)x \in \left(0, \frac{\pi}{4}\right):
1+sin⁡2x=∣sin⁡x+cos⁡x∣=sin⁡x+cos⁡x\sqrt{1 + \sin 2x} = |\sin x + \cos x| = \sin x + \cos x
 Step 2: Reduce and differentiate
sin⁡x+cos⁡xsin⁡x+cos⁡x=1  ⟹  ddx(1)=0\frac{\sin x + \cos x}{\sin x + \cos x} = 1 \implies \frac{d}{dx}(1) = 0
 Answer: (2)
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