The Curves That Refuse to Be Solved
Some equations tie $x$ and $y$ together so tightly that you cannot solve for $y$ at all. The unit circle $x^2 + y^2 = 1$ is one of them, and yet it clearly describes a shape and clearly has a slope at every point. An implicit function is how mathematics handles exactly these relationships.
What Is an Implicit Function?
An implicit function is a relationship between variables written as a single equation set equal to zero, $F(x, y) = 0$, in which $y$ is not isolated on one side. The equation still pairs $x$-values with $y$-values, but the dependence is hidden inside the equation rather than spelled out. Examples include:
$$x^2 + y^2 = 25, \qquad xy = 1, \qquad x^3 + y^3 = 6xy$$
Each of these defines $y$ in terms of $x$ implicitly. A function normally means one input gives one output, and some implicit equations bend that rule by pairing an $x$ with more than one $y$, which is why they are treated as relations that may define one or several functions on different pieces of their graph.
What Is the Difference Between Implicit and Explicit Functions?
An explicit function states the output directly: $y = f(x)$, with $y$ alone on the left. An implicit function leaves the two variables mixed together. The distinction is not about difficulty but about form.
Feature | Explicit function | Implicit function |
|---|---|---|
Form | $y = f(x)$ | $F(x, y) = 0$ |
$y$ isolated? | Yes | No |
Example | $y = 3x^2 + 2$ | $x^2 + y^2 = 25$ |
One $y$ per $x$? | Always | Not always |
Solve for $y$? | Already solved | Sometimes impossible |
The same relationship can often be written both ways. From $x + y = 4$ you can solve explicitly to get $y = 4 - x$. But from $x^3 + y^3 = 6xy$ there is no clean explicit form, so the implicit equation is the only practical description. A quick visual test helps: if a vertical line can cross the graph more than once, no single explicit $y = f(x)$ covers the whole curve.
How Do You Differentiate an Implicit Function?
Implicit differentiation finds the derivative $\frac{dy}{dx}$ without first solving for $y$. The trick is to treat $y$ as a function of $x$ and apply the chain rule every time you differentiate a term containing $y$. The steps are always the same:
Differentiate both sides of the equation with respect to $x$.
Apply the chain rule to every $y$-term, so $\frac{d}{dx}(y^n) = n y^{n-1}\frac{dy}{dx}$.
Collect all $\frac{dy}{dx}$ terms on one side.
Solve for $\frac{dy}{dx}$ by factoring and dividing.
Take the circle $x^2 + y^2 = 25$. Differentiate each term with respect to $x$:
$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$
$$2x + 2y\frac{dy}{dx} = 0$$
$$2y\frac{dy}{dx} = -2x$$
$$\frac{dy}{dx} = -\frac{x}{y}$$
The $\frac{dy}{dx}$ attached to the $y$-term is the chain rule doing its job, and it is exactly the piece beginners forget.
What Are the Properties of Implicit Functions?
A few features set implicit functions apart:
They may define several functions at once. The circle splits into an upper branch $y = \sqrt{25 - x^2}$ and a lower branch $y = -\sqrt{25 - x^2}$; the single implicit equation holds both.
They need not be expressible explicitly. Curves like $x^3 + y^3 = 6xy$ have no elementary explicit form, yet still have a well-defined slope everywhere the curve is smooth.
A local guarantee exists. The implicit function theorem says that near a point where the curve is smooth, $y$ can be treated as a genuine function of $x$, which is what makes implicit differentiation valid.
The derivative usually depends on both variables. For the circle, $\frac{dy}{dx} = -\frac{x}{y}$ involves $y$, so you need a point's full coordinates, not just its $x$-value, to get the slope.
Where Are Implicit Functions Used?
"What is the slope of a curve you cannot solve?"
That question shows up far beyond the textbook circle.
Conic sections and geometry. Circles, ellipses, and hyperbolas are all naturally implicit, and finding tangents and normals to them relies on implicit differentiation.
Related rates. When two changing quantities are locked by an equation, like the sides of a sliding ladder, implicit differentiation with respect to time links their rates without isolating either.
Economics. Indifference curves and production isoquants are level curves $F(x, y) = c$, and their slopes carry direct economic meaning.
Physics and chemistry. Laws such as the ideal gas relation tie several quantities together implicitly, so a change in one is read off the shared equation.
At Bhanzu, trainers first ask students to find $\frac{dy}{dx}$ for a circle two ways, once by solving for $y$ and once implicitly, so the shortcut proves itself before it is trusted.
Examples of Implicit Functions
The set runs from a clean circle, through the chain-rule trap, to product-rule terms and a tangent slope at a point.
Example 1
Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$.
Differentiate both sides with respect to $x$:
$$2x + 2y\frac{dy}{dx} = 0$$
$$2y\frac{dy}{dx} = -2x$$
$$\frac{dy}{dx} = -\frac{x}{y}$$
Final answer: $\frac{dy}{dx} = -\dfrac{x}{y}$.
Example 2
Find $\frac{dy}{dx}$ for $x^3 + y^3 = 9$.
Wrong attempt. A student differentiates term by term and writes $3x^2 + 3y^2 = 0$, treating $y^3$ as if $y$ were a constant. That gives $y^2 = -x^2$, which has no real solution, a signal that something broke. The missing piece is the chain rule: $y$ depends on $x$, so its derivative carries a $\frac{dy}{dx}$.
The fix. Differentiate correctly:
$$3x^2 + 3y^2\frac{dy}{dx} = 0$$
$$3y^2\frac{dy}{dx} = -3x^2$$
$$\frac{dy}{dx} = -\frac{x^2}{y^2}$$
Final answer: $\frac{dy}{dx} = -\dfrac{x^2}{y^2}$. Every $y$-term earns a $\frac{dy}{dx}$; drop it and the algebra collapses.
Example 3
Find $\frac{dy}{dx}$ for $xy = 1$.
The left side is a product, so use the product rule, and the $y$-factor still needs the chain rule:
$$x\frac{dy}{dx} + y = 0$$
$$x\frac{dy}{dx} = -y$$
$$\frac{dy}{dx} = -\frac{y}{x}$$
Final answer: $\frac{dy}{dx} = -\dfrac{y}{x}$.
Example 4
Find $\frac{dy}{dx}$ for $x^2 + xy + y^2 = 7$.
Differentiate each term, using the product rule on $xy$:
$$2x + \left(x\frac{dy}{dx} + y\right) + 2y\frac{dy}{dx} = 0$$
$$x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y$$
$$\frac{dy}{dx}(x + 2y) = -(2x + y)$$
$$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$$
Final answer: $\frac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$. The term students most often mishandle here is the mixed $xy$, which needs the product rule, not a single derivative.
Example 5
Find the slope of the tangent to $x^2 + y^2 = 25$ at the point $(3, 4)$.
From Example 1, $\frac{dy}{dx} = -\frac{x}{y}$. Substitute the point:
$$\frac{dy}{dx}\bigg|_{(3,4)} = -\frac{3}{4}$$
Final answer: the tangent slope at $(3, 4)$ is $-\dfrac{3}{4}$, matching the figure at the top of this article.
Example 6
Find $\frac{dy}{dx}$ for the folium of Descartes, $x^3 + y^3 = 6xy$.
Differentiate, using the product rule on the right side:
$$3x^2 + 3y^2\frac{dy}{dx} = 6\left(x\frac{dy}{dx} + y\right)$$
$$3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$$
$$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$$
$$\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$$
Final answer: $\frac{dy}{dx} = \dfrac{2y - x^2}{y^2 - 2x}$, a slope no explicit form could give as cleanly.
Common Mistakes
Mistake 1: Forgetting the chain-rule factor on y-terms
Where it slips in: Differentiating any term that contains $y$.
Don't do this: Write $\frac{d}{dx}(y^2) = 2y$, treating $y$ as if it were a constant.
The correct way: Because $y$ depends on $x$, the chain rule gives $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$. The memorizer who knows the power rule but forgets $y$ is a function of $x$ loses the whole factor and gets a broken answer.
Mistake 2: Skipping the product rule on mixed terms
Where it slips in: A term like $xy$ or $x^2y$ where both variables appear.
Don't do this: Differentiate $xy$ as $\frac{dy}{dx}$ alone, or as $1$.
The correct way: Treat $xy$ as a product: $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$. The rusher who moves fast often collapses the product into a single term and drops the $y$.
Mistake 3: Insisting on solving for y first
Where it slips in: Facing an equation like $x^3 + y^3 = 6xy$.
Don't do this: Burn time trying to isolate $y$ before differentiating, when no clean explicit form exists.
The correct way: Differentiate implicitly straight away; you never need $y$ alone. Recognising when an equation cannot be made explicit is half the skill.
Conclusion
An implicit function ties $x$ and $y$ together as $F(x, y) = 0$ rather than as $y = f(x)$.
Explicit and implicit are two forms of the same relationship, but some curves have only the implicit form.
Implicit differentiation finds $\frac{dy}{dx}$ by differentiating both sides and applying the chain rule to every $y$-term.
The most common error is dropping the $\frac{dy}{dx}$ factor on $y$-terms, followed by mishandling product terms like $xy$.
Implicit functions describe conics, related rates, indifference curves, and physical laws that no explicit formula captures cleanly.
To build fluency with functions and calculus alongside a teacher, explore Bhanzu's algebra tutor sessions, a high school math tutor, or flexible math tutoring. Want a trainer to walk your child through implicit differentiation step by step? Book a free demo class.
Read More
Vertical Line Test — the quick check for whether a curve is a function.
Zeros of a Function — reading solutions straight from $F(x) = 0$.
Even and Odd Function — symmetry properties that also apply to implicit curves.
Differentiation of Trigonometric Functions — the chain rule with trig terms.
Quadratic Equations — the implicit conics behind many curves.
Was this article helpful?
Your feedback helps us write better content