Qual è la derivata di csc ^ 2 (x) csc2(x)?
Qual è la derivata di csc ^ 2 (x) csc2(x)? Risposta: d/dx[csc^2(x)]= -2cotxcsc^2xddx[csc2(x)]=−2cotxcsc2x Spiegazione: csc^2(x)=1/sin^2(x)csc2(x)=1sin2(x) d/dx[csc^2(x)]=d/dx[1/sin^2(x)]ddx[csc2(x)]=ddx[1sin2(x)] d/dx[1/sin^2(x)]=d/dx[[sin(x)]^{-2}]ddx[1sin2(x)]=ddx[[sin(x)]−2] lasciare u=sinxu=sinx d/dx[[sin(x)]^{-2}]=d/{du}[u^{-2}]d/dx[sinx]ddx[[sin(x)]−2]=ddu[u−2]ddx[sinx] d/{du}[u^{-2}]= -2u^{-3}ddu[u−2]=−2u−3 d/dx[sinx] = cosxddx[sinx]=cosx d/dx[[sin(x)]^{-2}]=-2u^{-3}cosx=-{2cosx}/{sin^3x}ddx[[sin(x)]−2]=−2u−3cosx=−2cosxsin3x cosx/sinx=cotx => -{2cosx}/{sin^3x}=-{2cotx}/{sin^2x}cosxsinx=cotx⇒−2cosxsin3x=−2cotxsin2x 1/sin^2x=csc^2x => -2cotxcsc^2x 1sin2x=csc2x⇒−2cotxcsc2x d/dx[csc^2(x)]= -2cotxcsc^2xddx[csc2(x)]=−2cotxcsc2x