ln(1+sin2x^2)的导数怎么求?
问题描述:
ln(1+sin2x^2)的导数怎么求?
答
[ln(1+sin2x^2)]'
=1/(1+sin2x^2)*(1+sin2x^2)'
=cos2x^2/(1+sin2x^2)*(2x^2)'
=4xcos2x^2/(1+sin2x^2)
答
[ln(1+sin2x^2)]'
=1/[1+sin2x^2]*[1+sin2x^2]'
=1/[1+sin2x^2]*(sin2x^2)'
=1/[1+sin2x^2]*2sin2xcos2x*2
=4sin2xcos2x/[1+sin2x^2]
答
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)