Implicit Differentiation: The Chain Rule Trick

Most functions you meet in early calculus are written explicitly: you have $y$ isolated on one side, such as $y = 3x^2 + 1$. But many important curves cannot be rearranged into that tidy form. The equation of a circle, $x^2 + y^2 = 25$, defines a relationship between $x$ and $y$ without explicitly solving for either variable. To find slopes and tangent lines on such curves, we need a technique called implicit differentiation.

This guide walks through the concept from first principles, builds a repeatable step-by-step method, and then applies it to progressively harder worked examples. By the end you will be comfortable differentiating any implicitly defined curve and finding its tangent lines, second derivatives, and connections to related rates problems.

Why Ordinary Differentiation Falls Short

When a function is written as $y = f(x)$, you differentiate each term with respect to $x$ directly. The power rule, product rule, and chain rule all work straightforwardly because every term is expressed in terms of $x$.

Now consider the circle $x^2 + y^2 = 25$. You could solve for $y$ to get $y = \pm\sqrt{25 - x^2}$, but this splits the curve into two separate functions (the upper and lower semicircles) and introduces a square root that complicates further work. For an equation like $x^3 + y^3 = 6xy$ (the folium of Descartes), it is not possible to isolate $y$ algebraically at all. Implicit differentiation handles both cases elegantly.

The Core Idea: The Chain Rule in Disguise

Implicit differentiation rests on a single insight: treat $y$ as a function of $x$, even though you have not solved for it. Whenever you differentiate a term containing $y$, apply the chain rule. For instance:

The outer derivative gives $2y$ (power rule), and the chain rule multiplies by $\frac{dy}{dx}$ because $y$ itself depends on $x$. This factor of $\frac{dy}{dx}$ (often written $y'$) is what we ultimately solve for.

Step-by-Step Method

Follow these four steps for any implicitly defined equation:

1. Differentiate every term on both sides of the equation with respect to $x$. Apply the chain rule to every term that contains $y$, appending a factor of $\frac{dy}{dx}$.
2. Collect all terms containing $\frac{dy}{dx}$ on one side of the equation and move everything else to the other side.
3. Factor out $\frac{dy}{dx}$.
4. Divide both sides by the remaining factor to isolate $\frac{dy}{dx}$.

The answer will usually contain both $x$ and $y$. That is perfectly normal for implicit differentiation.

Worked Example 1: The Circle

Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$.

Step 1. Differentiate both sides with respect to $x$:

Step 2. Isolate the $\frac{dy}{dx}$ term:

Step 3. Divide both sides by $2y$:

This tells us that the slope at any point $(x, y)$ on the circle is $-x/y$. At $(3, 4)$, for example, the slope is $-3/4$, which matches geometric intuition: the radius to $(3,4)$ has slope $4/3$, and the tangent is perpendicular to the radius.

Worked Example 2: An Ellipse

Find $\frac{dy}{dx}$ for $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Step 1. Differentiate:

Step 2. Simplify and isolate:

At the point $\left(\frac{3\sqrt{2}}{2},\,\sqrt{2}\right)$ on the ellipse, the slope is:

Worked Example 3: The Folium of Descartes

Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$.

Step 1. Differentiate. The left side is straightforward. The right side requires the product rule because it contains both $x$ and $y$:

Step 2. Collect $\frac{dy}{dx}$ terms on one side:

Step 3. Factor and divide:

The curve passes through $(3, 3)$. At that point:

So the tangent line at $(3, 3)$ has slope $-1$, meaning it slopes downward at 45 degrees.

Finding the Equation of a Tangent Line

Once you have $\frac{dy}{dx}$ evaluated at a point $(x_0, y_0)$, the tangent line equation is:

Example. For the circle $x^2 + y^2 = 25$ at $(3, 4)$, we found $m = -3/4$. The tangent line is:

Worked Example 4: A Curve Requiring the Product Rule

Find $\frac{dy}{dx}$ for $x^2 y + xy^2 = 6$.

Differentiate term by term. For $x^2 y$, use the product rule: $2xy + x^2 \frac{dy}{dx}$. For $xy^2$, again use the product rule: $y^2 + 2xy\,\frac{dy}{dx}$. The right side gives 0.

Collect and factor:

At the point $(1, 2)$ (verify: $1 \cdot 2 + 1 \cdot 4 = 6$):

Second Derivatives by Implicit Differentiation

You can find $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ again with respect to $x$, remembering that $y$ (and therefore $\frac{dy}{dx}$) still depends on $x$.

Example. For $x^2 + y^2 = 25$, we found $\frac{dy}{dx} = -\frac{x}{y}$. Differentiate using the quotient rule:

Multiply numerator and denominator by $y$:

The last simplification uses $x^2 + y^2 = 25$. Notice that the second derivative is always negative when $y > 0$ (top half of the circle), confirming that the curve is concave down there, exactly as we would expect.

Connection to Related Rates

Related rates problems are implicit differentiation in disguise. Instead of differentiating with respect to $x$, you differentiate with respect to time $t$. Every variable that changes over time picks up a $\frac{d}{dt}$ factor via the chain rule.

Example. A spherical balloon is being inflated so that its volume increases at $100\,\text{cm}^3/\text{s}$. How fast is the radius changing when $r = 10\,\text{cm}$?

Start with the volume formula:

Differentiate both sides with respect to $t$:

Substitute $\frac{dV}{dt} = 100$ and $r = 10$:

The structure is identical to implicit differentiation: apply the chain rule to every variable, then solve for the unknown rate.

Common Mistakes to Avoid

• Forgetting the chain rule factor. Every time you differentiate a $y$ term, you must multiply by $\frac{dy}{dx}$. Omitting this is the most common error.
• Dropping the product rule. Terms like $xy$ or $x^2 y$ require the product rule because they contain two variable quantities.
• Substituting the point too early. Always find the general expression for $\frac{dy}{dx}$ first, then substitute the coordinates.
• Sign errors when rearranging. Be meticulous when moving terms across the equals sign.

When to Use Implicit Differentiation

Use implicit differentiation whenever the equation relating $x$ and $y$ is difficult or impossible to solve for $y$ explicitly. Common scenarios include:

• Circles, ellipses, and other conic sections
• Higher-degree polynomial curves (e.g., the folium of Descartes)
• Equations involving transcendental functions of $y$, such as $e^y + x = y$
• Related rates problems where multiple quantities change with time
• Implicit equations arising from level curves of multivariable functions

Frequently Asked Questions

What is the difference between explicit and implicit differentiation?

Explicit differentiation works on functions written as $y = f(x)$, where $y$ is isolated on one side. Implicit differentiation works on equations where $x$ and $y$ are intertwined, such as $x^2 + y^2 = 25$. The technique is the same (differentiation rules), but implicit differentiation adds the chain rule factor $\frac{dy}{dx}$ wherever $y$ appears.

Why does implicit differentiation give an answer containing both x and y?

Because $y$ was never isolated as a function of $x$, the derivative naturally depends on both variables. To evaluate the slope at a particular point, you substitute both the $x$ and $y$ coordinates of that point into the expression.

Can I use implicit differentiation on any equation?

Yes, as long as the equation defines $y$ as a differentiable function of $x$ (at least locally). At points where $\frac{dy}{dx}$ would involve division by zero, the tangent line is vertical and the derivative does not exist in the usual sense.

How does implicit differentiation relate to multivariable calculus?

If $F(x, y) = 0$ defines the curve, then the implicit function theorem gives $\frac{dy}{dx} = -\frac{F_x}{F_y}$, where $F_x$ and $F_y$ are the partial derivatives. This is exactly the formula you derive by implicit differentiation, confirming the connection between single-variable and multivariable viewpoints.

Is implicit differentiation on the AP Calculus exam?

Yes. Implicit differentiation is a standard topic on both the AP Calculus AB and BC exams. You should be able to find first and second derivatives implicitly, determine tangent line equations, and solve related rates problems that rely on the same technique.