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

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(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

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

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) = $

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