y=asinx =>dy/dx = acosx
z=cosx => dz/dx = -sinx
---------
u = e^(ax) . (y-z)/(a^2+1)
du/dx
=[1/(a^2+1)] . [ a(y-z) + dy/dx - dz/dx ] .e^(ax)
=[1/(a^2+1)] . [ a(y-z)+ acosx + sinx ] .e^(ax)
=[1/(a^2+1)] . [ a( asinx -cosx)+ acosx + sinx ] .e^(ax)
=[ sinx/(a^2+1)] . ( a^2 + 1) .e^(ax)
= sinx .e^(ax)追问
z=cosx => dz/dx = -sinx
---------
u = e^(ax) . (y-z)/(a^2+1)
du/dx
=[1/(a^2+1)] . [ a(y-z) + dy/dx - dz/dx ] .e^(ax)
=[1/(a^2+1)] . [ a(y-z)+ acosx + sinx ] .e^(ax)
=[1/(a^2+1)] . [ a( asinx -cosx)+ acosx + sinx ] .e^(ax)
=[ sinx/(a^2+1)] . ( a^2 + 1) .e^(ax)
= sinx .e^(ax)追问
倒数第三行是怎么化简到倒数第二行的？
温馨提示：答案为网友推荐，仅供参考
相似回答