Qual è la derivata di # csc ^ 2 (x) #?
Risposta:
#d/dx[csc^2(x)]= -2cotxcsc^2x#
Spiegazione:
#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}]#
lasciare #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#