Question

$\frac{d}{dx}\left(x^2\sin \left(\frac{1}{x}\right)\right)=$

• A. $\cos \left(\frac{1}{x}\right)+2x\sin \left(\frac{1}{x}\right)$
• B. $2x\sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)$
• C. $\cos \left(\frac{1}{x}\right)-2x\sin \left(\frac{1}{x}\right)$
• D. none of these

Answer 1

The correct option is B

$2x\sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)$

Explanation for correct option:

Given, $\frac{d}{dx}\left(x^2\sin \left(\frac{1}{x}\right)\right)$

Apply theorem for multiplication.$\frac{d}{dx}\left(u·v\right)=u\frac{dv}{dx}+v\frac{du}{dx}$

$$\begin{gathered} =2x\sin \left(\frac{1}{x}\right)+x^2\cos \left(\frac{1}{x}\right)\times \left(-\frac{1}{x^2}\right) \\ =2x\sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right) \end{gathered}$$

Hence, correct answer is option $\left(B\right)$