ln(1+sin2x^2)的导数怎么求?

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ln(1+sin2x^2)的导数怎么求?ln(1+sin2x^2)的导数怎么求?ln(1+sin2x^2)的导数怎么求?因为(lnx)''=1/x所以[ln(1+sin2x^2)]''=[1/(1+sin

ln(1+sin2x^2)的导数怎么求?
ln(1+sin2x^2)的导数怎么求?

ln(1+sin2x^2)的导数怎么求?
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)

[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]

[ln(1+sin2x^2)]'
=1/(1+sin2x^2)*(1+sin2x^2)'
=cos2x^2/(1+sin2x^2)*(2x^2)'
=4xcos2x^2/(1+sin2x^2)