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Algebra of Derivatives

If y = sin ...

Question

If $y = sin x [\frac{1}{sin x \cdot sin 2 x} + \frac{1}{sin 2 x sin 3 x} + \dots$ $+ \frac{1}{sin n x sin (n + 1) x}]$ then $\frac{d y}{d x} =$

A

cot x - cot (n + 1) x

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B

(n + 1) cos e c^{2} (n + 1) x - cos e c^{2} x

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C

{csc}^{2} x - (n + 1) cos e c^{2} (n + 1) x

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D

cot x + cot (n + 1) x

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Solution

The correct option is C $(n + 1) cos e c^{2} (n + 1) x - cos e c^{2} x$
$\begin{matrix} y = [\frac{1}{sin x sin 2 x} + \frac{1}{sin 2 x sin 3 x} + . . . . . . . . . . . + \frac{1}{sin n x sin (n + 1) x}] y = [\frac{sin x}{sin x sin 2 x} + \frac{sin x}{sin 2 x sin 3 x} + . . . . . . . . . . . + \frac{sin x}{sin n x sin (n + 1) x}] y = [\frac{sin (2 x - x)}{sin 2 x sin x} + \frac{sin (3 x - 2 x)}{sin 3 x sin 2 x} + . . . . . . . . . . . + \frac{sin (n + 1) x - n x}{sin (n + 1) x sin n x}] y = [\frac{sin (2 x) cos x - cos 2 x sin x}{sin 2 x sin x} + \frac{sin (3 x) cos 2 x - cos 3 x sin 2 x}{sin 3 x sin 2 x} + . . . . . . . . . . . + \frac{sin (n + 1) x cos n x - cos (n + 1) x sin x}{sin (n + 1) x sin n x}] y = [\frac{sin (2 x) cos x}{sin 2 x sin x} - \frac{cos 2 x sin x}{sin 2 x sin x} + \frac{sin (3 x) cos 2 x}{sin 2 x sin x} \frac{- cos 3 x sin 2 x}{sin 3 x sin 2 x} + . . . . . . . . . . . + \frac{sin (n + 1) x cos n x}{sin (n + 1) x sin n x} - \frac{cos (n + 1) x sin x}{sin (n + 1) x sin n x}] y = [\frac{cos x}{sin x} - \frac{c o s 2 x}{sin 2 x} + \frac{c o s 2 x}{sin 2 x} - \frac{c o s 3 x}{sin 3 x} + . . . . . + \frac{cos n x}{sin n x} - \frac{cos (n + 1) x}{sin (n + 1) x}] y = [cot x - cot 2 x + cot 2 x - cot 3 x + . . . cot n x - cot (n + 1) x] y = [cot x - cot (n + 1) x] \frac{d y}{d x} = - cos e c^{2} x - (- cos e c^{2} (n + 1) x) (n + 1) \frac{d y}{d x} = (n + 1) cos e c^{2} (n + 1) x - cos e c^{2} x \end{matrix}$

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