来自冉维丽的问题
求导y=ln(tan(x/2))
求导y=ln(tan(x/2))
1回答
2020-05-20 09:28
求导y=ln(tan(x/2))
求导y=ln(tan(x/2))
y'=1/(tan(x/2))*(tan(x/2))'=1/(tan(x/2))*(sec^2(x/2))*(x/2)'=1/(2sin(x/2)*cos(x/2))=1/sin(x)=csc(x)