What is the derivative of $- \frac{1}{{sin (x)}^{2}}$ $-1/(sin(x)^2)$ ?

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What is the derivative of $- \frac{1}{{sin (x)}^{2}}$ $-1/(sin(x)^2)$ ?

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Answer 1 · 2016-12-26T18:21:49

$\frac{d}{d x} (- \frac{1}{{sin}^{2} x}) = \frac{2 cos x}{{sin}^{3} x}$ $d/(dx) (-1/sin^2x) = (2cosx)/(sin^3x)$

Explanation:

Based on the chain rule:

$\frac{d}{d x} (- \frac{1}{{sin}^{x}}) = \frac{d}{d sin x} (- \frac{1}{{sin}^{2} x}) \cdot \frac{d sin x}{d x} = \frac{2}{{sin}^{3} x} cos x$ $d/(dx) (-1/sin^x) = d/(dsinx) (-1/sin^2x) * (dsinx)/(dx)= 2/(sin^3x)cosx$

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What is the derivative of $- \frac{1}{{sin (x)}^{2}}$ $-1/(sin(x)^2)$ ?

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