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need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 7

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Answer: 2 \sin (4 x+2)

Hint: You must know the rules of solving derivation of trigonometric function.

Given:  \sin ^{2}(2 x+1)


Let  y=\sin ^{2}(2 x+1)

Differentiating with respect to x,

\frac{d y}{d x}=\frac{d}{d x}\left[\sin ^{2}(2 x+1)\right]

\frac{d y}{d x}=2 \sin (2 x+1) \frac{d}{d x} \sin (2 x+1)          [ using chain rule ]

\frac{d y}{d x}=2 \sin (2 x+1) \cos (2 x+1) \frac{d}{d x}(2 x+1)            \left[\frac{d}{d x} \sin x=\cos x\right]

\begin{aligned} &\frac{d y}{d x}=4 \sin (2 x+1) \cos (2 x+1) \\ &\frac{d y}{d x}=2 \sin (4 x+2) \end{aligned}                [\therefore \sin 2 A=2 \sin A \cos A]

\frac{d y}{d x}=2 \sin (4 x+2)      

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