frac{d}{dx} left( x^2 sin frac{1}{x} right) = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To find the derivative of $x^2 \sin \frac{1}{x}$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u = x^2$ and

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) To find the derivative of $x^2 \sin \frac{1}{x}$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u = x^2$ and $v = \sin \frac{1}{x}$.Then $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = \cos \left( \frac{1}{x} \right) \cdot \frac{d}{dx} \left( \frac{1}{x} \right) = \cos \left( \frac{1}{x} \right) \cdot \left( -\frac{1}{x^2} \right)$.Applying the product rule:$\frac{d}{dx} \left( x^2 \sin \frac{1}{x} \right) = x^2 \left( -\frac{1}{x^2} \cos \frac{1}{x} \right) + \sin \left( \frac{1}{x} \right) (2x)$$= -\cos \left( \frac{1}{x} \right) + 2x \sin \left( \frac{1}{x} \right)$$= 2x \sin \left( \frac{1}{x} \right) - \cos \left( \frac{1}{x} \right)$.

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

Similar Questions

If $f(2) = 4$ and $f'(2) = 1$,then $\mathop {\lim }\limits_{x \to 2} \frac{xf(2) - 2f(x)}{x - 2} = $

The differential coefficient of ${x^6}$ with respect to ${x^3}$ is

If $y=\frac{\cos x}{1+\sin x}$,then
$(a)$ $\frac{dy}{dx}=\frac{-1}{1+\sin x}$
$(b)$ $\frac{dy}{dx}=\frac{1}{1+\sin x}$
$(c)$ $\frac{dy}{dx}=-\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$
$(d)$ $\frac{dy}{dx}=\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$

$\frac{d}{dx} \left( x^3 \tan^2 \frac{x}{2} \right) =$

If $f(x) = \frac{\sin^2 x}{1+\cot x} + \frac{\cos^2 x}{1+\tan x}$,then the value of $f^{\prime}\left(\frac{\pi}{6}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module