If y = sin x cdot sin 2x cdot sin 3x cdot ldo | vedclass

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(B) Given,$y = \sin x \cdot \sin 2x \cdot \sin 3x \cdot \ldots \cdot \sin nx$.Taking the natural logarithm on both sides:$\ln y = \ln(\sin x) + \ln(\s

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$y \cdot \sum_{k=1}^{n} k \cot kx$

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(B) Given,$y = \sin x \cdot \sin 2x \cdot \sin 3x \cdot \ldots \cdot \sin nx$.Taking the natural logarithm on both sides:$\ln y = \ln(\sin x) + \ln(\sin 2x) + \ln(\sin 3x) + \ldots + \ln(\sin nx)$Differentiating both sides with respect to $x$ using the chain rule:$\frac{1}{y} \cdot y^{\prime} = \frac{d}{dx}(\ln \sin x) + \frac{d}{dx}(\ln \sin 2x) + \ldots + \frac{d}{dx}(\ln \sin nx)$$\frac{y^{\prime}}{y} = \frac{\cos x}{\sin x} + 2 \cdot \frac{\cos 2x}{\sin 2x} + 3 \cdot \frac{\cos 3x}{\sin 3x} + \ldots + n \cdot \frac{\cos nx}{\sin nx}$$\frac{y^{\prime}}{y} = \sum_{k=1}^{n} k \cot kx$Therefore,$y^{\prime} = y \cdot \sum_{k=1}^{n} k \cot kx$.

If $y = \sin x \cdot \sin 2x \cdot \sin 3x \cdot \ldots \cdot \sin nx$,then $y^{\prime}$ is

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