Question

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

• A. $\sin \left({\log }_{e}x\right)$
• B. $\cos \left({\log }_{e}x\right)$
• C. $y^2$
• D. $-y$

Answer 1

The correct option is D

$-y$

Explanation for the correct option:

$y=\sin \left({\log }_{e}x\right)$

Step 1: Differentiating with respect to $x$ both sides :

$$\begin{gathered} \frac{dy}{dx}=\cos \left({\log }_{e}x\right)\times \frac{1}{x} \\ =\frac{\cos \left({\log }_{e}x\right)}{x} \end{gathered}$$

So,$\frac{dy}{dx}=\frac{\cos \left({\log }_{e}x\right)}{x}$

Step 2: Using Quotient rule and again differentiating with respect to $x$ both sides :

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{x\left(-\sin \left({\log }_{e}x\right)\right)\times {\displaystyle \frac{1}{x}}-\cos \left({\log }_{e}x\right)}{x^2} \\ =\frac{-\cos \left({\log }_{e}x\right)-\sin \left({\log }_{e}x\right)}{x^2} \end{gathered}$$

So, $\frac{d^{2}y}{d{x}^2}=\frac{-\cos \left({\log }_{e}x\right)-\sin \left({\log }_{e}x\right)}{x^2}$

$$\begin{gathered} x^2\frac{d^{2}y}{d{x}^2}=-\cos \left({\log }_{e}x\right)-\sin \left({\log }_{e}x\right) \\ =-x\frac{dy}{dx}-y \end{gathered}$$

$\Rightarrow x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=-y$

Hence, Option (D) is the correct answer.