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

Question

65725 views around the world

You can reuse this answer
Creative Commons License

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#