Over the next few screens we explore implicit differentiation. We’ll develop an easy-to-use, can’t-fail approach that you can apply to any homework or exam problem.

Recall *the* most important thing about “functions” that you’ve been taught since you first learned about them: **for every input value, a function returns one and only one output value.** That’s why the “vertical line test” works: if any vertical line passes through more than one *y* -value on a curve, then then at that value of *x* the curve has more than one associated *y*-value, and so that curve does *not* represent a function.

As a quick example, the graph on the left below shows the familiar function $f(x) = x^3 + 1.$ You can drag the vertical line along the *x*-axis anywhere you’d like. No matter where you put it, that line intersects the curve in only *one* place: each *x*-value has a *single* associated *y*-value, the quality which makes this a function.

By contrast, the graph on the right shows the familiar unit circle, described by the relation $x^2 + y^2 = 1.$ Notice that the vertical line intersects the circle in *two* places everywhere, except at the ends. For instance, at $x = 0,$ we have the two associated values $y = -1$ and $y = 1.$ This relationship is thus *not* a function.

Graph of the function *$f(x) = x^3 + 1$* versus *x*

Graph of $x^2 + y^2 = 1$, which is NOT a function

We’re going to look now at two different *ways* that we can define a function.

A function is **explicitly defined** if it’s given in the form $f(x) =\, \rule{0.6cm}{0.15mm},$ or similar. For instance, these are all explicitly defined functions:

\[f(x) = x^3 + 1 \qquad \sin(\theta) = g(\theta ) \qquad y(t) = e^t – \ln(t)\]
You might think of a particular explicit function definition as an instruction list, with *explicit* directions for how to get the output $f(x),$ $g(\theta),$ or $y(t)$ from the input to each. (For *f:* Take the input *x*, cube it, and then add 1.” And so forth.) In each case, for every input value you obtain one, and only one, output value, as you must since these are all functions.

By contrast, a function is **implicitly defined** if you’re given an equation that *relates* values of *x* and *y,* say without providing explicit instructions for how to find either one. (Actually, we may implicitly define more than one function from a single equation, a key point we’ll come back to shortly.) These equations all define functions $y(x)$ implicitly:

\[y – x^3 -1 = 0 \quad\qquad x^2 + y^2 = 1 \quad\qquad (x^2-y)^2 = x^2-y^2 \]

Let’s take a brief look at each of these.

Equation 1: $y – x^3 – 1 = 0$

You may have recognized that equation as being the same as, after a tiny bit of algebraic manipulation, the equation $y = f(x) = x^3 + 1$ that we graphed above. While above we defined the equation explicitly, the fact that we have to do *any* manipulation means that the function $y(x)$ is defined *implicitly* by the equation $y – x^3 – 1 = 0.$ It just so happens that in *this* case we can easily solve the given equation to find the single function $y(x) = x^3 + 1$ that “works” to make the original equation valid. As a check, if we substitute the function we’ve found back into the original equation, we get

\begin{align*}

\text{Original equation:} \\[8px]
y -x^3 -1 &= 0 \\[8px]
\text{Let’s substitute } y = f(x) = x^3 + 1: \\[8px]
\left[x^3 + 1 \right] – x^3 – 1 &\overbrace{=}^? 0 \\[8px]
0 &= 0 \quad \cmark

\end{align*}

This last bit may seem silly, but it illustrates a key point that will be more important as we continue onward:

A function

Things get a little more interesting with our next example:

Equation 2: $x^2 + y^2 = 1$

This is of course our familiar unit circle, which we also graphed above, calling specific attention to the fact that this equation is *not* a function. In this case you certainly can (and almost surely have) solved the equation for *y:* $y = \pm \sqrt{1 – x^2}.$ What this really means is that the circle equation *implicitly* defines (at least) *two* unique functions:

\[y_1(x) = \sqrt{1 – x^2} \quad \text{and} \quad y_2(x) = -\sqrt{1 – x^2} \]

You probably recognize $y_1(x)$ and $y_2(x)$ as the top and bottom halves of the circle respectively. Notice that both $y_1$ and $y_2$ are indeed functions: each returns a single value of *y* for a given input value *x.*

As before, you can verify that this function is implicitly defined by the circle equation by substituting $y = f(x)$ for both pieces of the function, to again obtain the identity “1 = 1.”

Let’s consider our third equation.

Equation 3: $(x^2-y)^2 = x^2-y^2$

Desmos will certainly plot this equation for you, as shown. You might then even recognize the curve as the upside-down logo of the company Meta, formerly Facebook. It’s also a specific instance of what’s known as a Lissajous curve.

But looking at the equation $(x^2-y)^2 = x^2-y^2$, it’s not immediately clear that we can solve it for one or more functions $y(x) = \dots$ as we did above. (In fact you can, but it’s not immediately apparent – and more importantly, it actually doesn’t matter. And we’ll see on the next screen many equations that you simply cannot solve for $y(x).$) In reality, such equations arise in Nature, and hence we use them in Mathematics and Physics, and often there is no way to solve for an explicit definition of $y(x).$ For better or worse, most of the equations you’ve encountered up until now have been formulated precisely so that you *can* solve them for the dependent variable; indeed, that’s what a large part your algebra classes have been all about. But Nature certainly isn’t formulated to only make use of such equations, and hence you’ve perhaps been misled into thinking we can always write an explicit definition for functions to describe anything we care to. This simply isn’t so.

Solve the Lissajous equation for $y(x)$

As we noted above, it’s not immediately apparent that you can solve the equation $(x^2-y)^2 = x^2-y^2$ for $y(x).$ As it turns out, you can:

\begin{align*}

(x^2-y)^2 &= x^2-y^2 \\[8px] x^4 – 2x^2 y + y^2 &= x^2-y^2 \\[8px] 2y^2 – 2x^2 y + (x^4-x^2) &= 0

\end{align*}

Using the quadratic formula where $a=2$, $b=-2x^2$, and $c=(x^4-x^2)$, we can find an equation for $y(x)$ after doing a lot of simplifying.

\begin{align*}

y(x) &= \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\[8px] &= \frac{-(-2x^2) \pm \sqrt{(-2x^2)^2 – 4(2)(x^4-x^2)}}{2(2)} \\[8px] &= \frac{2x^2 \pm \sqrt{4x^4 – 8x^4 + 8x^2}}{4} \\[8px] &= \frac{2x^2 \pm \sqrt{8x^2 – 4x^4}}{4} \\[8px] &= \frac{2x^2 \pm \sqrt{4x^2(2 – x^2)}}{4} \\[8px] &= \frac{2x^2 \pm 2|x|\sqrt{(2 – x^2)}}{4} \\[8px] &= \frac{x^2 \pm |x|\sqrt{(2 – x^2)}}{2}

\end{align*}

Hence we can find an expression for $y(x)$ that satisfies the original equation. But really it doesn’t matter; the key point is that $y(x)$ is implicitly defined by $(x^2-y)^2 = x^2-y^2,$ which is enough to know that we can find $y'(x)$ … as we will on the next screen!

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Here’s the thing: we don’t *have* to have an explicit definition for *y* in order to be able to work with such an equation in Calculus. And the next screen we’ll see how we can take its derivative, and hence determine the slope of the curve at any point we choose.

There is a theorem, known aptly enough as the “Implicit Function Theorem,” that provides conditions under which a function is defined implicitly by a given equation and is differentiable. The theorem is, however, beyond our ability to discuss in beginning Calculus since it requires understanding and tools that we won’t develop until much later.

Hence, we — along with every other website, textbook, online homework question, … — ask that you *assume* that if you’re given an equation in terms of *x* and *y,* then that equation does indeed determine a differentiable function *f* (or differentiable functions $f_1,$ $f_2,$ …). You can thus safely use the tools we’ll develop on the next screen without having to prove anything else first.

We’ve tried to be careful on this screen to say that we are either *explicitly defining* a function or *implicitly defining* a function or functions.

Essentially everywhere else in the world, and on this site, you will see the words “implicit function.” For instance, you will routinely see the equation $x^2 + y^2 = 1$” called “an implicit function.” This is actually sloppy terminology, but we all use it . . . even though it causes students confusion when first learning this topic, because it makes it seem like there’s a new *kind* of function, one that (maybe??) has more than one output value *y* for every input value *x?!?* But NO: a function does indeed have only a *single* output value for each input value. That is a fundamental definition that does not change.

Here’s what you need to keep in mind:

“an equation that implicitly defines one or more functions.”

For example, as we saw above, when anyone says “the implicit function $x^2 + y^2 = 1,$” what they really mean is that the equation defines (at least) the two functions $y_1(x) = \sqrt{1 – x^2}$ and $y_2(x) = -\sqrt{1 – x^2}.$

Let’s conclude this screen with some quick Check Questions.

For each equation below, decide whether $y(x)$ is implicitly defined, explicitly defined, or undefined.

- A function has
*one and only one*output value for every input value in its domain. - A function is explicitly defined if it is in the form $y(x) =\, \rule{0.6cm}{0.15mm},$ or $f(x) =\, \rule{0.6cm}{0.15mm},$ or similar. No manipulation is required to isolate the output value on one side of the definition.
- A function is, or functions are, implicitly defined if the given equation
*relates*the values of*x*and*y,*without providing explicit instructions for how to find either one. - When we, or anyone, says “this implicit function,” what they really mean is “this function is (or functions are) implicitly defined by the given equation.”

On the next screen, we’ll see how we can easily find the derivative of an implicitly defined function, using a technique known (aptly enough) as “implicit differentiation.”

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