来自高仁祥的问题
y=lncotx/2求导y=lncot(x/2)求导,
y=lncotx/2求导
y=lncot(x/2)求导,
1回答
2020-05-22 06:59
y=lncotx/2求导y=lncot(x/2)求导,
y=lncotx/2求导
y=lncot(x/2)求导,
y'=1/cot(x/2)*[-csc²(x/2)]*1/2
=-1/2*1/sin²(x/2)*sin(x/2)/cos(x/2)
=-1/[2sin(x/2)cos(x/2)]
=-1/sinx
=-cscx