Question

How do you implicitly differentiate `-1=y^2+(xy-e^y)/(x)`?

Answer 1

The key to answering implicit differentiation problems is to derive portions of the equation involving a `y` the same as you would the portions with `x`, remembering to multiply by a `dy/dx`, which you will later isolate to solve.

Explanation:

Rewriting the problem,

`-1 = y^2 + (xy)/x - (e^y)/x`
`-1 = y^2 + y - (e^y)/x`

We now derive both sides of the equation, remembering to include a `dy/dx` when deriving any functions of `y`:

`d/dx( -1 ) = d/dx( y^2 + y - e^y x^-1 )`
`0 = 2y dy/dx + dy/dx - (e^y dy/dx x^-1 + (-1)e^y x^-2)`
`0 = 2y dy/dx + dy/dx - e^y x^-1 dy/dx + e^y x^-2`

We now move all terms with `dy/dx` to one side of the equations in order to isolate the term:

`2y dy/dx + dy/dx - e^y x^-1 dy/dx = -e^y x^-2`

We can now isolate `dy/dx`:

`dy/dx(2y + 1 - e^y x^-1) = -e^y x^-2`
`dy/dx = (-e^y x^-2)/(2y + 1 - e^y x^-1)`

The `x` terms with negative powers can now be rewritten to simplify:

`dy/dx = ((-e^y)/x^2)/(2y + 1 - e^y / x)`
`dy/dx = (-e^y)/(2yx^2 + x^2 - e^y x)`
`dy/dx = (e^y)/(e^y x - x^2 - 2yx^2)`

Giving us the simplified answer:

`dy/dx = (e^y) / (x(e^y - x[1 + 2y])`