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Implicit
Differentiation

Learn to differentiate equations where \(y\) isn't isolated: find \(\frac{dy}{dx}\) for curves like circles, ellipses, and beyond.

AP Calculus AB

What Is an Implicit Equation?

Most equations you have seen are explicit: \(y\) is isolated on one side, like \(y = x^2 + 3x\). But many important curves cannot be written this way.

Explicit
\(y = f(x)\) : output is directly expressed in terms of input.
Implicit
\(F(x,y) = 0\) : \(x\) and \(y\) are tangled together in one equation.

Why We Need Implicit Differentiation

When a curve is defined implicitly, solving for \(y\) can be difficult or impossible. Implicit differentiation lets us find \(\frac{dy}{dx}\) directly from the original equation, without ever isolating \(y\).

x y (0.6, 0.8)
\(x^2 + y^2 = 1\)
 tangent line

Curves You Can't Solve for \(y\)

Many real-world curves are only expressible implicitly. For these, implicit differentiation is essential, not optional.

Folium of Descartes
\(x^3 + y^3 = 6xy\)
Elliptic Curve
\(y^2 = x^3 - x + 1\)
Astroid
\(x^{2/3} + y^{2/3} = 4\)

The Core Idea

Treat \(y\) as a function of \(x\), even if you can't write it explicitly. Every time you differentiate a term containing \(y\), apply the chain rule and multiply by \(\frac{dy}{dx}\).

\(\frac{d}{dx}[y^2]\)
\(= 2y \cdot \frac{dy}{dx}\)
\(\frac{d}{dx}[\sin y]\)
\(= \cos y \cdot \frac{dy}{dx}\)
\(\frac{d}{dx}[e^y]\)
\(= e^y \cdot \frac{dy}{dx}\)

Applying the Derivative Operator

The core technique is to apply the derivative operator, \(\frac{d}{dx}\), to both sides of the implicit equation. Think of it as performing the same operation on both sides to maintain the balance.

We then differentiate term-by-term. The derivative of a constant (like 25) is zero. For terms with \(y\), we must remember to apply the chain rule.

Why the \(\frac{dy}{dx}\) Factor Appears

\(x\)
\(y(x)\)
\(f(y)\)
\(\frac{dy}{dx}\)
\(f'(y)\)

When we differentiate a function of \(y\) with respect to \(x\), we're linking rates of change. A change in \(x\) causes \(y\) to change (at a rate of \(\frac{dy}{dx}\)), which in turn causes \(f(y)\) to change (at a rate of \(f'(y)\)).

Explicit vs Implicit Differentiation

Explicit Implicit
Starting form \(y = f(x)\) \(F(x,y) = 0\)
Method Direct rules (power, product, etc.) Differentiate both sides; chain rule on \(y\)
Result \(\frac{dy}{dx}\) in terms of \(x\) only \(\frac{dy}{dx}\) may involve both \(x\) and \(y\)
When to use \(y\) is easily isolated \(y\) is tangled with \(x\) or unsolvable

Step-by-Step Method

1
Differentiate both sides of the equation with respect to \(x\).
2
Apply the chain rule: whenever you differentiate a term containing \(y\), multiply by \(\frac{dy}{dx}\).
3
Use the product/quotient rule as needed for mixed terms like \(xy\) or \(\frac{x}{y}\).
4
Collect all \(\frac{dy}{dx}\) terms on one side of the equation.
5
Factor out \(\frac{dy}{dx}\) and solve by dividing.

Example 1: The Circle

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

1
Differentiate both sides: \(\frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25]\)
2
Apply rules: \(2x + 2y \cdot \frac{dy}{dx} = 0\)
3
Solve: \(2y \cdot \frac{dy}{dx} = -2x\)
4
Result: \(\dfrac{dy}{dx} = -\dfrac{x}{y}\)

Example 2: Product Rule Meets Implicit

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

1
Differentiate \(x^2 y\) using the product rule: \(2xy + x^2 \frac{dy}{dx}\)
2
Differentiate \(xy^2\) using the product rule: \(y^2 + x \cdot 2y \frac{dy}{dx}\)
3
Full equation: \(2xy + x^2 \frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0\)
4
Factor: \(\frac{dy}{dx}(x^2 + 2xy) = -(2xy + y^2)\)
5
Solve: \(\dfrac{dy}{dx} = -\dfrac{2xy + y^2}{x^2 + 2xy}\)

Example 3: Trigonometric Implicit Equations

Find \(\frac{dy}{dx}\) for \(\sin(xy) = x\).

1
Differentiate: \(\cos(xy) \cdot \frac{d}{dx}[xy] = 1\)
2
Chain + product rule: \(\cos(xy)\!\left(y + x\frac{dy}{dx}\right) = 1\)
3
Expand: \(y\cos(xy) + x\cos(xy)\frac{dy}{dx} = 1\)
4
Solve: \(\dfrac{dy}{dx} = \dfrac{1 - y\cos(xy)}{x\cos(xy)}\)

Example 4: Finding a Tangent Line

Find the equation of the tangent line to \(x^3 + y^3 = 9\) at the point \((1, 2)\).

1
Differentiate: \(3x^2 + 3y^2 \frac{dy}{dx} = 0\)
2
Solve: \(\frac{dy}{dx} = -\frac{x^2}{y^2}\)
3
Evaluate at \((1,2)\): \(\frac{dy}{dx} = -\frac{1^2}{2^2} = -\frac{1}{4}\)
4
Tangent line: \(y - 2 = -\frac{1}{4}(x - 1)\) or \(y = -\frac{1}{4}x + \frac{9}{4}\)

Going Further: Second Derivatives

You can find \(\frac{d^2y}{dx^2}\) by differentiating \(\frac{dy}{dx}\) implicitly again, then substituting the first derivative result.

The second derivative tells us about concavity. Since \(-\frac{25}{y^3}\) is negative when \(y > 0\), the upper semicircle is concave down, which matches the geometry of a circle.

Common Mistakes

✕ Forgetting \(\frac{dy}{dx}\)
Writing \(\frac{d}{dx}[y^2] = 2y\) instead of \(2y \cdot \frac{dy}{dx}\). Every \(y\)-term needs the chain rule factor.
✕ Skipping the Product Rule
Writing \(\frac{d}{dx}[xy] = \frac{dy}{dx}\) instead of \(y + x\frac{dy}{dx}\). Mixed terms need the product rule.
✕ Wrong Collection
Failing to gather all \(\frac{dy}{dx}\) terms on one side before factoring. This leads to algebraic errors in the final answer.

Summary of Key Formulas

Expression Implicit Derivative
\(y^n\) \(ny^{n-1}\frac{dy}{dx}\)
\(xy\) \(y + x\frac{dy}{dx}\)
\(\sin y\) \(\cos y \cdot \frac{dy}{dx}\)
\(e^y\) \(e^y \cdot \frac{dy}{dx}\)
\(\ln y\) \(\frac{1}{y}\cdot\frac{dy}{dx}\)
\(x^2 + y^2 = r^2\) \(\frac{dy}{dx} = -\frac{x}{y}\)

Question 1 of 5

What is \(\frac{dy}{dx}\) for the equation \(x^2 + y^2 = 16\)?

Question 2 of 5

Using implicit differentiation, what is \(\frac{d}{dx}[y^3]\)?

Question 3 of 5

For the equation \(xy = 10\), what is \(\frac{dy}{dx}\)?

Question 4 of 5

What is the slope of the tangent line to \(x^2 + xy + y^2 = 7\) at the point (1, 2)?

Question 5 of 5

Given \(\sin y + y = x^2\), which expression equals \(\frac{dy}{dx}\)?