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(A) Given $y = x \sin x$. Taking the natural logarithm on both sides,we get $\ln y = \ln(x \sin x)$. Using the property $\ln(ab) = \ln a + \ln b$,we h

Vedclass · Accepted Answer

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \cot x$

Vedclass · Answer

(A) Given $y = x \sin x$. Taking the natural logarithm on both sides,we get $\ln y = \ln(x \sin x)$. Using the property $\ln(ab) = \ln a + \ln b$,we have $\ln y = \ln x + \ln(\sin x)$. Differentiating both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln(\sin x))$. $\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \frac{1}{\sin x} \cdot \cos x$. $\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \cot x$. Thus,the correct option is $A$.

If $y = x \sin x$,then

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