Question

求导的数学题

Answer 1

f(x)= 2x.(cosx)^2
f'(x)
=2 [ x(2cosx).(-sinx) + (cosx)^2 ]
=2[ -xsin2x + (cosx)^2 ]
f'(μ)=2[ -μsin2μ + (cosμ)^2 ]

追问

你好我想问问我画出来的地方是怎么来的

谢谢

追答

y=f(u) , u=sin(x^2)

y=f(sin(x^2))

y' 

= f'(sin(x^2)) . d/dx(sin(x^2))

= f'(sin(x^2)) . (-cos(x^2) .(2x) )

=-2x.cos(x^2).f'(sin(x^2))

y''

=-2[ x.cos(x^2). d/dx{ f'(sin(x^2))} + f'(sin(x^2)) .d/dx { x.cos(x^2) } ]

//

来自这里！

d/dx { x.cos(x^2) }

= x [-sin(x^2)](2x) + cos(x^2)