求复合导数.y=lnsin(1/x)y=lncot(x/2)要过程

Question

求复合导数.
y=lnsin(1/x)
y=lncot(x/2)
要过程

Answer 1

答案分别为-(1/x²)cot(1/x)和-cscx
解法:
我用导数符号dy/dx了,因为用这个的链式法则求复合函数的导数时很好用的
y=lnsin(1/x)
dy/dx=dlnsin(1/x)/dsin(1/x)·dsin(1/x)/d(1/x)·d(1/x)/dx
=1/sin(1/x)·cos(1/x)·(-1/x²)
=-(1/x²)cot(1/x)
y=lncot(x/2)
dy/dx=dlncot(x/2)/dcot(x/2)·dcot(x/2)/d(x/2)·d(x/2)/dx
=1/cot(x/2)·[-csc²(x/2)]·(1/2)
=-(1/2)sin(x/2)/cos(x/2)·1/sin²(x/2)
=-(1/2)·1/[sin(x/2)cos(x/2)]
=-1/sinx
=-cscx