Question

2xcosx^2的导数？

2xcos*2

Answer 1

若是 [2x(cosx)^2]' = 2[(cosx)^2 + 2xcosx(-sinx)] = 2[(cosx)^2 - xsin2x]
若是 [2xcos(x^2)]' = 2[(cos(x^2) - 4x^2sin(x^2)]

Answer 2

y=2x.cos(x^2)
y'
=2{ x.[-sin(x^2)].(x^2)' + cos(x^2) .(x)']
=2{ x.[-sin(x^2)].(2x) + cos(x^2) .(1)]
=2[-2x^2.sin(x^2) + cos(x^2)]