Question

How do you use implicit differentiation to find dy/dx given `xe^y-y=5`?

Answer 1

`dy/dx = (e^y)/(1-xe^y)`

Explanation:

When we differentiate `y` wrt `x` we get `dy/dx`.

However, we cannot differentiate a non implicit function of `y` wrt `x`. But if we apply the chain rule we can differentiate a function of `y` wrt `y` but we must also multiply the result by `dy/dx`.

When this is done in situ it is known as implicit differentiation.

We have:

`xe^y-y=5`

Differentiate wrt `x` (applying product rule):

`(x)(e^ydy/dx)+(1)(e^y) - dy/dx=0`
`:. xe^ydy/dx + e^y - dy/dx=0`
`:. e^y = dy/dx-xe^ydy/dx`
`:. (1-xe^y)dy/dx = e^y`
`:. dy/dx = e^y/(1-xe^y)`

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find `y` explicitly as a function of `x`, only implicitly through the equation `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`. Therefore we can write `F(x, y) = 0` as `F(x, y(x)) = 0`. Differentiating both sides of this, using the partial chain rule gives us

`(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))`

So Let `F(x,y) = xe^y-y-5`; Then;

`(partial F)/(partial x) = e^y`

`(partial F)/(partial y) = xe^y-1`

And so:

`dy/dx = -(e^y)/(xe^y-1) = (e^y)/(1-xe^y)`, as before.