Uses of implicit differentiation

In your example, y is a function of x. x is also a function of y. If you pick a value for x, there is a unique y that satisfies the equality. If you pick a value for y, there a unique x that satisfies the equality. Let's find y as a function of x

    x^2y = 5 => y = 5x^-2 <------ this is y(x)

That's all it means. This a function, because as mentioned, x uniquely determines y.

Now let's look at an example where y is not a function of x - the standard example of a circle of radius r:

    x^2 + y^2 = r^2

solving for y gives

    y = +- sqrt(r^2 - x^2).

This is NOT a function, because for a value of x, we don't know which value of y to choose - it could be the positive case or the negative case. Geometrically, this means a line (like y = c) can intersect the circle in two places, so trying to find the single intersection point is not well defined.

But the gradient y' IS a function, a function of both x and y and we can find it by implicit differentiation (with respect to x):

    2x + 2yy' = 0 => y' = -x/y  <---- this is y'(x,y)

Geometrically this means that given a point (x,y) on a circle, the gradient is a well defined quantity, even though y(x) is not.

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Also: I had a look at your submission history and I really don't think that HN is the right place for most of your questions. I'm just procrastinating by answering this. You should probably be using math.stackexchange.com or a similar site.

This is a perfect example of how math is very simple. It just isn't always simple in a way our brains are wired to naturally follow. With the result that when people "get" a concept, they beat themselves up for not getting something so simple.

Implicit differentiation is simple in that way.

Let's take your example. xxy = 5. This implicitly defines some function y. xxy is an expression. And 5 is another expression. And the two are equal for all x. Now note that you can differentiate any expression you want to. For instance the derivative of xxy is 2xy + xxy'. And the derivative of 5 is 0.

The entire idea of implicit differentiation is this. If two expressions are the same for all x, then they have the same tangent lines, and therefore the same derivatives. Thus you don't need to solve for y first. You can just differentiate and solve for y' instead. This is convenient when y is difficult to solve for.

In this case you find that 2xy + xxy' = 0 so xxy' = -2xy and so y' = -2y/x.

That's all there is to it.

Now two notes of interest. The first is that regular differentiation is really just a special case of implicit differentiation. We get y=f(x) and differentiate both sides to get y'=f'(x). The second is that equations like y' = -2y/x are the start of a very important subject called differential equations.

Your question is a little vague, but I'll try.

What it is: Differentiate both sides of an equation. That's it.

Why it works: When two things are equal, if we transform them in identical ways (being careful about what "identical" means), then the results are equal. So if two differentiable functions are equal, then their derivatives must be equal.

What it's good for: In spite of all the carefully selected examples we're given in calculus class, the fact is that solving equations is often very difficult, and usually impossible. So if someone writes some "f(x,y) = k", and asks you to solve for y in terms of x, well, in general, you can't do it.

Now, suppose you want to find dy/dx in the above situation. The "usual" way is to solve for y and then differentiate. But what if you cannot do that? If you can find the derivative, w.r.t. x, of f(x,y), then you can apply implicit differentiation, to get something like g(x,y)*dy/dx + h(x,y) = 0. Then finding dy/dx is easy.

So: What it is good for is finding a derivative while avoiding the pain (or the impossibility) of solving an equation.

Of course, that leads to the question of what derivatives are good for, but I'm hoping you have some inkling of the answer to that one. If not, then back up, and learn more about derivatives.