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)`
`[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.
We substitute our expression for `y’` from above, `dy/dx = (-y)/(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:
The answer looks quite different, but it does have the same value.
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