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

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Answer 1 · 2018-04-13T20:14:40

$d/dx[csc^2(x)]= -2cotxcsc^2x$

$csc^2(x)=1/sin^2(x)$

$d/dx[csc^2(x)]=d/dx[1/sin^2(x)]$

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

let $u=sinx$

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

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

$d/dx[sinx] = cosx$

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

$cosx/sinx=cotx => -{2cosx}/{sin^3x}=-{2cotx}/{sin^2x}$

$1/sin^2x=csc^2x => -2cotxcsc^2x$

$d/dx[csc^2(x)]= -2cotxcsc^2x$

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