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

Question

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

Aicharya · Accepted Answer

(b) Given, $y=\sin \left(\log \left(x^2+2 x+1ight)ight)$ $ \Rightarrow \quad y=\sin [2 \log (x+1)] $ So, $\frac{d y}{d x}=\left[\cos (2 \log (x+1)] 	imes\left(\frac{2}{x+1}ight)ight.$ $ \Rightarrow \quad(x+1) \frac{d y}{d x}=2 \cos (2(\log (x+1)) $ On differentiating both side with respect to ' $x$ ' $ \begin{aligned} & (x+1) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=-2(\sin (2 \log (1+x)))\left(\frac{2}{x+1}ight) \ & \Rightarrow \quad(x+1)^2 \frac{d^2 y}{d x^2}+(x+1) \frac{d y}{d x}=-4 y \end{aligned} $

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

Solution