Question

ln(1+sin2x^2)的导数怎么求?

Answer 1

因为(lnx)'=1/x

所以[ln(1+sin2x^2)]' = [1/(1+sin2x^2)]*(1+sin2x^2)'

=[1/(1+sin2x^2)]*(sin2x^2)'

=[1/(1+sin2x^2)]*(cos2x^2)*(2x^2)'

= [cos2x^2/(1+sin2x^2)]*4x

= 4xcos2x^2/(1+sin2x^2)