Question

求此函数导数

Answer 1

y=ln[tan(x/2)]
dy/dx = d{ln[tan(x/2)]} / dx
= 1/[tan(x/2)] * d[tan(x/2)] / dx
= 1/[tan(x/2)] * 1/cos²(x/2) *d(x/2) / dx
= 1/[tan(x/2)] * 1/cos²(x/2) * 1/2
= 1/{2sin(x/2)cos(x/2)}
= 1/sinx

Answer 2

解：
y'
=cot(x/2)·sec²(x/2)·（1/2）
=(1/2)cot(x/2)·sec²(x/2)
=1/[2sin(x/2)cos(x/2)]
=1/sinx
=cscx