First derivative:

Now xy is a product, so we use Product Formula to obtain:


And we learned in the last section on Implicit Differentiation that


We can write this as:

`d/dx(y^2) = 2yy’`

Putting it together, here is the first derivative of our implicit function:

`xy’ + y + 2yy’ = 0`

[I am using `y’` instead of `dy/dx`. It is easier to type and quite easy to read.]

We could write this as: `dy/dx = -y/(x+2y)`

Second derivative:

`[xy” + y’] + [ y’] +` ` [2yy” + y’(2y’)] ` `= 0`

and this simplifies to:

`(x + 2y)y” + 2y’ + 2(y’)^2 = 0`

We can solve for `y”`:

`y” = [-2(y’ + (y’)^2)] / (x + 2y)`

We could leave our answer as is, or simplify it even further by re-expressing it in terms of `x` and `y` only, as follows.

Further simplification

We substitute our expression for `y’` from above, `dy/dx = (-y)/(x+2y)`:

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


Now multiply by `(x+2y)^2/(x+2y)^2` to give






Here's a movie giving a different view of this example. It uses:

  1. `dy/dx` notation;
  2. A different approach to the problem (in the movie, we find the expression for `dy/dx` first, then differentiate that to get the second derivative). The result is a simpler expression for the second derivative.

The answer looks quite different, but it does have the same value.

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