begin{aligned} & y=sin left(log left(x^2+2 x+ | vedclass

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(B) Given,$y = \sin(\log(x^2 + 2x + 1))$.Since $x^2 + 2x + 1 = (x+1)^2$,we have $y = \sin(2 \log(x+1))$.Differentiating with respect to $x$:$\frac{dy}

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$-4 y$

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(B) Given,$y = \sin(\log(x^2 + 2x + 1))$.Since $x^2 + 2x + 1 = (x+1)^2$,we have $y = \sin(2 \log(x+1))$.Differentiating with respect to $x$:$\frac{dy}{dx} = \cos(2 \log(x+1)) 	imes \frac{2}{x+1}$.Multiplying by $(x+1)$:$(x+1) \frac{dy}{dx} = 2 \cos(2 \log(x+1))$.Differentiating again with respect to $x$ using the product rule:$(x+1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = -2 \sin(2 \log(x+1)) 	imes \frac{2}{x+1}$.Multiplying the entire equation by $(x+1)$:$(x+1)^2 \frac{d^2y}{dx^2} + (x+1) \frac{dy}{dx} = -4 \sin(2 \log(x+1))$.Since $y = \sin(2 \log(x+1))$,we get:$(x+1)^2 \frac{d^2y}{dx^2} + (x+1) \frac{dy}{dx} = -4y$.

$\begin{aligned} & y=\sin \left(\log \left(x^2+2 x+1\right)\right) \\ & \Rightarrow(x+1)^2 \frac{d^2 y}{d x^2}+(x+1) \frac{d y}{d x}= \end{aligned}$

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