三角函數的微分

h(x) = sin2x cos2x

h'(x)

= sin2x (cos2x)' + cos2x (sin2x)'

= sin2x (-2sin2x) + cos2x (2cos2x)

= 2[(cos2x)^2 - (sin2x)^2]

= 2cos4x

h(x)=Sin2xCos2x=(1/2)Sin4x
h'(x)=(1/2)*4Cos4x=2Cos4x

先講微分的積法則
[f(x)g(x)]'=f'(x)g(x)=f(x)+g'(x)

還有鏈鎖律
[f(g(x))]'=f'(g(x))*g'(x)

h(x)=sin2xcos2x
=(sin2x)(cos2x)
h'(x)=(sin2x)'cos(2x)+(sin2x)(cos2x)'

再來 sin2x的微分
令f(x)=sinx
g(x)=2x
sin2x=f(g(x))
[f(g(x))]'=f'(g(x))*g'(x)
=cos(2x)*2
=2cos2x

同理
cos(2x)=-2sin2x

h'(x)=(sin2x)'cos(2x)+(sin2x)(cos2x)'
=(2cos2x)*(cos2x)+(sin2x)(-2sin2x)
=2(cos2x)^2-2(sin2x)^2

希望沒有錯誤