Tiff

Qual é a derivada de `csc ^ 2 (x)`? Responda: `d/dx[csc^2(x)]= -2cotxcsc^2x` Explicação: `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}]` deixar `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`