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JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1
If y=sin⁡x⋅sin⁡2x⋅sin⁡3x…⋅sin⁡nxy=\sin x \cdot \sin 2 x \cdot \sin 3 x \ldots \cdot \sin n x, then dydx=……\frac{d y}{d x}=\ldots \ldots
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Correct option: B
y∑k=1nkcot⁡kxy \sum\limits_{{k=1}}^n k \cot k x
Option B (y∑k=1nkcot⁡kxy \sum\limits_ {{k=1}}^n k \cot k x) is correct because by taking the logarithm of both sides and differentiating, we can express dy/dx in terms of y and a sum of cotangents.

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