Question

F(x)=f(cos~2x)+f(sin~2x)的导数 急————————

Answer 1

F(x)=f(cos²x)+f(sin²x),f 可导,令 u=cos²x,v=sin²x,则

du/dx = 2 cosx * (-sinx) = - sin(2x),dv/dx = 2 sinx * cosx = sin(2x)

F' (x) = f '(u) * du/dx + f '(v) * dv/dx

= sin(2x) * [ - f '(u) + f '(v) ]

= .