Question

How do you implicitly differentiate `ln(x/y)=x-y`?

Answer 1

It is

#d(lnx-lny)/dx=d(x-y)/dx=> 1/x-[(y')/y]=1-dy/dx=> dy/dx-dy/dx*[1/y]=1-1/x=> dy/dx=[1-1/x]/[1-1/y]#