Question

What is the implicit derivative of `1= e^(xy-y)-xy`?

Answer 1

`dy/dx=(ye^(xy-y)-y)/(x-xe^(xy-y)+e^(xy-y) )`

Explanation:

`1=e^(xy-y)-xy`

`0=ye^(xy-y)+(x-1)e^(xy-y) dy/dx-y-xdy/dx`

`xdy/dx-(x-1)e^(xy-y) dy/dx=ye^(xy-y)-y`

`dy/dx(x-(x-1)e^(xy-y) )=ye^(xy-y)-y`

`dy/dx(x-xe^(xy-y)+e^(xy-y) )=ye^(xy-y)-y`

`dy/dx=(ye^(xy-y)-y)/(x-xe^(xy-y)+e^(xy-y) )`