Assertion (A): frac{d}{d x}left(frac{x^2 sin | vedclass

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(C) Assertion $(A)$: Let $y = \frac{x^2 \sin x}{\log x}$. Taking $\log$ on both sides,$\log y = \log(x^2) + \log(\sin x) - \log(\log x) = 2 \log x + \

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$A$ is true,$R$ is false

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(C) Assertion $(A)$: Let $y = \frac{x^2 \sin x}{\log x}$. Taking $\log$ on both sides,$\log y = \log(x^2) + \log(\sin x) - \log(\log x) = 2 \log x + \log(\sin x) - \log(\log x)$. Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{2}{x} + \frac{\cos x}{\sin x} - \frac{1}{\log x} \cdot \frac{1}{x} = \frac{2}{x} + \cot x - \frac{1}{x \log x}$. Thus,$\frac{dy}{dx} = y \left(\cot x + \frac{2}{x} - \frac{1}{x \log x}\right) = \frac{x^2 \sin x}{\log x} \left(\cot x + \frac{2}{x} - \frac{1}{x \log x}\right)$. So,$A$ is true. Reason $(R)$: Using logarithmic differentiation,$\frac{d}{dx} \left(\frac{uv}{w}\right) = \frac{uv}{w} \frac{d}{dx} (\log u + \log v - \log w) = \frac{uv}{w} \left(\frac{u'}{u} + \frac{v'}{v} - \frac{w'}{w}\right)$. The given formula in $R$ has a plus sign instead of a minus sign for the $w$ term,so $R$ is false.

Assertion $(A)$: $\frac{d}{d x}\left(\frac{x^2 \sin x}{\log x}\right)=\frac{x^2 \sin x}{\log x} \left(\cot x+\frac{2}{x}-\frac{1}{x \log x}\right)$
Reason $(R)$: $\frac{d}{d x}\left(\frac{u v}{w}\right)=\frac{u v}{w}\left[\frac{u^{\prime}}{u}+\frac{v^{\prime}}{v}-\frac{w^{\prime}}{w}\right]$

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Assertion $(A)$: $\frac{d}{d x}\left(\frac{x^2 \sin x}{\log x}\right)=\frac{x^2 \sin x}{\log x} \left(\cot x+\frac{2}{x}-\frac{1}{x \log x}\right)$ Reason $(R)$: $\frac{d}{d x}\left(\frac{u v}{w}\right)=\frac{u v}{w}\left[\frac{u^{\prime}}{u}+\frac{v^{\prime}}{v}-\frac{w^{\prime}}{w}\right]$