What is the derivative of ${csc}^{2} (x)$ $csc^2(x)$ ?

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What is the derivative of ${csc}^{2} (x)$ $csc^2(x)$ ?

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Answer 1 · 2018-04-13T20:14:40

$\frac{d}{d x} [{csc}^{2} (x)] = - 2 cot x {csc}^{2} x$ $d/dx[csc^2(x)]= -2cotxcsc^2x$

Explanation:

${csc}^{2} (x) = \frac{1}{{sin}^{2} (x)}$ $csc^2(x)=1/sin^2(x)$

$\frac{d}{d x} [{csc}^{2} (x)] = \frac{d}{d x} [\frac{1}{{sin}^{2} (x)}]$ $d/dx[csc^2(x)]=d/dx[1/sin^2(x)]$

$\frac{d}{d x} [\frac{1}{{sin}^{2} (x)}] = \frac{d}{d x} [{[sin (x)]}^{- 2}]$ $d/dx[1/sin^2(x)]=d/dx[[sin(x)]^{-2}]$

let $u = sin x$ $u=sinx$

$\frac{d}{d x} [{[sin (x)]}^{- 2}] = \frac{d}{d u} [u^{- 2}] \frac{d}{d x} [sin x]$ $d/dx[[sin(x)]^{-2}]=d/{du}[u^{-2}]d/dx[sinx]$

$\frac{d}{d u} [u^{- 2}] = - 2 u^{- 3}$ $d/{du}[u^{-2}]= -2u^{-3}$

$\frac{d}{d x} [sin x] = cos x$ $d/dx[sinx] = cosx$

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

$\frac{cos x}{sin x} = cot x \Rightarrow - \frac{2 cos x}{{sin}^{3} x} = - \frac{2 cot x}{{sin}^{2} x}$ $cosx/sinx=cotx => -{2cosx}/{sin^3x}=-{2cotx}/{sin^2x}$

$\frac{1}{{sin}^{2} x} = {csc}^{2} x \Rightarrow - 2 cot x {csc}^{2} x$ $1/sin^2x=csc^2x => -2cotxcsc^2x$

$\frac{d}{d x} [{csc}^{2} (x)] = - 2 cot x {csc}^{2} x$ $d/dx[csc^2(x)]= -2cotxcsc^2x$

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What is the derivative of ${csc}^{2} (x)$ $csc^2(x)$ ?

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