Calculus – Differentiating Equations Impossible to Solve for a Variable

Question

I thought I had a good idea on why/how implicit differentiation works until I read the following passage in my Calculus book:

Furthermore, implicit differentiation works just as easily for equations such as
$$x^5+5x^4y^2+3xy^3+y^5=1$$
which are actually impossible to solve for $y$ in terms of $x$

My problem with it is the following:

The way we go about differentiating, for instance, $xy=1$ is by differentiating the whole equation through with respect to $x$ and treating $y$ as a function of $x$. But (and at least that's how I see it) we can only treat $y$ as $f(x)$ because the equation determines $y$ as a function of $x$ in a relation that can ben written expliclity (in this case, $y=\frac{1}{x}$). If we have an equation such as the quoted one, in which we just can't solve for $y$, doesn't that mean that $y$ is not a function of $x$? In such a case, wouldn't treating it as such be an invalid move?

I hope I have made myself understood. Any clarification will be appreciated. Thanks

Answer 1

At a basic level, I think this question is really about the difference between saying that something exists, on the one hand, and being able to write a formula for it on the other. It's important to distinguish between three different (but closely related) ideas:

1. It may be that the relationship between $x$ and $y$ does not define $y$ as a function of $x$ because there is more than one $y$-value associated to a given $x$-value. To take a simple example, the equation of a unit circle ($x^2+y^2=1$) does not define $y$ as a function of $x$.
2. However, even if $y$ is not globally a function of $x$, it is nevertheless possible that locally (i.e. in the vicinity of some point) there may be a function of $x$ that "matches" the graph of the relationship, in a precise sense. For example, the point $(0.6, -0.8)$ lies on the bottom half of the unit circle. and the function $f(x) = -\sqrt{1-x^2}$ is a local solution for $y$ in terms of $x$ that includes that point. The Implicit Function Theorem provides conditions under which such a function exists.
3. On the other hand, even when such a local function exists, it may be impossible to write down an explicit formula for it. That, I think, is what your textbook means by "impossible to solve". It's not that the function doesn't exist, but rather that there is no way to write down a formula for the function.

To elaborate on this last point, consider the implicitly defined relation $$ y^3 + 2^y = \cos(2\pi x^2) + x $$ This relationship implicitly defines $y$ as a function of $x$: choose any specific value of $x$, say $x=2$. Then the right-hand side of the equation is $\cos(2\pi\cdot4) + 2 = 3$. The equation then asks us to find a value of $y$ such that $y^3 + 2^y = 3$. Such a $y$ is guaranteed to exist, and is in fact unique, as you can convince yourself of by looking at the graph of the function $h(t) = t^3 + 2^t$ (it's strictly increasing because its derivative is always positive, and its range is $(-\infty ,\infty)$ . And there's nothing special about the choice $x=2$ in this example; choose any value of $x$, and there is a unique $y$ value associated to that value of $x$ by the relation $ y^3 + 2^y = \cos(2\pi x^2) + x $. In fact you can see the graph of this implicitly-defined function below.

But go ahead, try to find a formula for explicitly calculating $y$ in terms of $x$. I'll wait.

(Okay, this is the point where someone jumps into the comments and says "Well, actually..." and goes on to explain that you can explicitly calculate $y$ in terms of $x$ by introducing a Lambert function or something. Let me try to pre-empt that by arguing that such a"solution" just sweeps the implicitness under the rug. In any case it misses the point of the example, which is that a relationship may implicitly define a function even if you lack an explicit formula for computing one variable in terms of the other.)

On the other hand, consider this closely-related example: $$ y^2 + 2^y = \cos(2\pi x^2) + x $$ (The only change is the exponent on the $y$ on the left-hand side.) This relationship most definitely does not define $y$ as a function of $x$, as can be seen in the graph below:

We can see that for many values of $x$ there are two different $y$ values that both satisfy $ y^2 + 2^y = \cos(2\pi x^2) + x $, so this is not a function. Nevertheless if we choose a point on the graph — $(2,1)$ is a convenient one — and zoom in on a neighborhood of that point, it locally looks like a function:

The power of implicit differentiation as a technique is precisely that it allows us to find an explicit formula for the slope of the tangent line at $(x,y)$, even when we can't find an explicit formula for $y$ in terms of $x$, and even when the "function" isn't really a function at all. (The trade-off is that the "explicit" formula for the slope is expressed in terms of both variables, so there is still some lurking implicitness in the problem.)