Question

What is the implicit derivative of `4= (x+y)^2`?

Answer 1

You can use calculus and spend a few minutes on this problem or you can use algebra and spend a few seconds, but either way you'll get `dy/dx=-1`.

Explanation:

Begin by taking the derivative with respect to both sides:
`d/dx(4)=d/dx(x+y)^2`

On the left, we have the derivative of a constant - which is just `0`. That breaks the problem down to:
`0=d/dx(x+y)^2`

To evaluate `d/dx(x+y)^2`, we need to use the power rule and the chain rule:
`d/dx(x+y)^2=(x+y)'*2(x+y)^(2-1)`
Note: we multiply by `(x+y)'` because the chain rule tells us we have to multiply the derivative of the whole function (in this case `(x+y)^2` by the inside function (in this case `(x+y)`).
`d/dx(x+y)^2=(x+y)'*2(x+y)`

As for `(x+y)'`, notice that we can use the sum rule to break it into `x'+y'`. `x'` is simply `1`, and because we don't actually know what `y` is, we have to leave `y'` as `dy/dx`:
`d/dx(x+y)^2=(1+dy/dx)(2(x+y))`

Now that we've found our derivative, the problem is:
`0=(1+dy/dx)(2(x+y))`

Doing some algebra to isolate `dy/dx`, we see:
`0=(1+dy/dx)(2x+2y)`
`0=2x+dy/dx2x+dy/dx2y+2y`
`0=x+dy/dxx+dy/dxy+y`
`-x-y=dy/dxx+dy/dxy`
`-x-y=dy/dx(x+y)`
`dy/dx=(-x-y)/(x+y)`

Interestingly, this equals `-1` for all `x` and `y` (except when `x=-y`). Therefore, `dy/dx=-1`. We could have actually figured this out without using any calculus at all! Look at the equation `4=(x+y)^2`. Take the square root of both sides to get `+-2=x+y`. Now subtract `x` from both sides, and we have `y=+-2-x`. Remember these from algebra? The slope of this line is `-1`, and since the derivative is the slope, we could have just said `dy/dx=-1` and avoided all that work.