设y=ln(sinx2+1),求dy

Question

如题所述

Answer 1

解
y'
=1/(sinx²+1)×(sinx²+1)'
=1/(sinx²+1)×[cosx²(x²)'+0]
=1/(sinx²+1)×2xcosx²
=2xcosx²/(sinx²+1)
dy=[2xcosx²/(sinx²+1)dx本回答被网友采纳