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ǰλãҳ >> ΪʲôCos(x+h)%Cosx=%2sin(x+h/2)sin(h/2) >>

ΪʲôCos(x+h)%Cosx=%2sin(x+h/2)sin(h/2)

Ͳʽ Уa=x+hb=x

Ϊcos(x+h)-cosx=-2sin(x+h/2)sin(h/2) :lim [cos(x+h)-cosx]/h =lim[-2sin(x+h/2)sin(h/2)/h] =lim[-sin(x+h/2)sin(h/2)/(h/2)] =lim[-sin(x+h/2)] =-sinx

ʽh0ʱļcosx. ͣ lim{h-->0}2sin(h/2)/h =lim{h-->0}sin(h/2)/(h/2) h/2=tʽ =lim{t-->0}sint/t ǻޣ =1. lim{h-->0}cos(x+h/2 =cosx ԭʽh-->0ʱΪcosx*1=cosx. ...

ǺͲʽ sin(x+h)-sinx=2cos(x +h/2)sinh/2 h0ʱlim[sin(x+h)-sinx]/h=lim2cos(x +h/2)sinh/2 /h ע⵽h0ʱlimcos(x +h/2)=cosx limsinh/2 /h=lim1/2 sinh/2 /(h/2)=1/2 lim[sin(x+h)-sinx]/h=cosx

Ϊд,涼ʡԵlim(h0) cosxĶ(-,+),cosxcos(x+h)ж [cos(x+h)-cosx]/h =-2sin[(x+h+x)/2]sin[(x+h-x)/2]/h =-sin[(2x+h)/2]*sin(h/2)/(h/2) =-sinx*1 =-sinx

֤f(x)=sin x ĵ cos x f'(x)=lim(h-->0)[sin(x+h)-sinx]/h =lim(h-->0)[2cos(x+h/2)sin(h/2)]/h =lim(h-->0)cos(x+h/2)*lim(h-->0)sin(h/2)/(h/2) =cosx lim(h-->0)sin(h/2)/(h/2)=1 f(x)=cos x ĵ -sin x f'(x)=lim(h...

h(x)= (-1-> sinx ) ( cos(t^2) +t ) dt h'(x)= cosx . { cos([sinx]^2) + sinx }

sin(A+B)=sinAcosB+cosAsinB (1) sin(A-B)=sinAcosB-cosAsinB (2) (1)-(2) 2cosAsinB = sin(A+B)- sin(A-B) now A+B=x+h (3)...

ǺĺͲʽ:sinx-siny=2cos[(x+y)/2]sin[(x-y)/2] ɵsin(x+h)-sinx=2cos[(x+h+x)/2]sin[(x+h-x)/2]=2cos...

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