Question

If $y=\sin \left(\ln x\right)$, then $x^2\frac{{\mathrm{d}}^{2}y}{d{x}^2}+x\frac{dy}{dx}$ is equal to

• A. $\sin \left(\ln x\right)$
• B. $\cos \left(\ln x\right)$
• C. $y^2$
• D. $-y$

Answer 1

The correct option is D

$-y$

The explanation for the correct option:

The given equation is $y=\sin \left(\ln x\right)$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \therefore \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left[\sin \left(\ln x\right)\right] \\ \Rightarrow \frac{dy}{dx}=\cos \left(\ln x\right)·\frac{d}{dx}\left(\ln x\right) \\ \Rightarrow \frac{dy}{dx}=\cos \left(\ln x\right)·\frac{1}{x} \\ \Rightarrow x\frac{dy}{dx}=\cos \left(\ln x\right) \end{gathered}$$

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(x\frac{dy}{dx}\right)=\frac{d}{dx}\left[\cos \left(\ln x\right)\right] \\ \Rightarrow \frac{dx}{dx}·\frac{dy}{dx}+x\frac{{\mathrm{d}}^{2}y}{d{x}^2}=-\sin \left(\ln x\right)·\frac{d}{dx}\left(\ln x\right)\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{u}\mathbf{v}\mathbf{\right)}\mathbf{=}\mathit{v}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{u}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\right] \\ \Rightarrow \frac{dy}{dx}+x\frac{{\mathrm{d}}^{2}y}{d{x}^2}=-\sin \left(\ln x\right)·\frac{d}{dx}\left(\ln x\right) \\ \Rightarrow \frac{dy}{dx}+x\frac{{\mathrm{d}}^{2}y}{d{x}^2}=-\sin \left(\ln x\right)·\frac{1}{x} \\ \Rightarrow x\frac{dy}{dx}+x^2\frac{{\mathrm{d}}^{2}y}{d{x}^2}=-\sin \left(\ln x\right) \\ \Rightarrow x^2\frac{{\mathrm{d}}^{2}y}{d{x}^2}+x\frac{dy}{dx}=-y\left[\because y=\sin \left(\mathrm{lnx}\right)\right] \end{gathered}$$

Therefore, $x^2\frac{{\mathrm{d}}^{2}y}{d{x}^2}+x\frac{dy}{dx}$ is equal to $-y$.

Hence, (D) is the correct option.