来自裴扬的问题
ln(1+sin2x^2)的导数怎么求?
ln(1+sin2x^2)的导数怎么求?
1回答
2020-02-07 00:27
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)