Implicit Differentiation: Study Guide (Calculus AB 3.2)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so derivatives are written in a compact form:

• dy/dx is the derivative of y with respect to x. Some books write y′; both are accepted.
• d/dx [ … ] means “differentiate the bracket with respect to x”.
• e^y means e raised to the power y; y³ means y cubed.

Explicit and implicit equations

An equation such as y = x³ − 2x is explicit: y is on its own, written as a formula in x. You already know how to differentiate it.

An equation such as x² + y² = 169 (a circle of radius 13) is implicit: x and y are mixed together. Near most points, part of the curve is still the graph of a function y(x), so it has a slope. But you may not be able, or want, to solve for y.

For this circle you can solve: the top half is y = √(169 − x²). The chain rule gives dy/dx = −x/√(169 − x²), and at x = 5 that is −5/12. The bottom half needs a second formula, y = −√(169 − x²).

Implicit differentiation finds both slopes at once, without solving.

The key idea: y is a function of x

Treat y as an unknown function of x, y = y(x). Then any expression in y is a composite: an outer function applied to the inner function y(x). The chain rule from Topic 3.1 says: differentiate the outer function, then multiply by the derivative of the inside. The derivative of the inside is dy/dx.

Expression	Rule used	d/dx of it
x³	power rule	3x²
y³	chain rule (outer u³, inner y)	3y² · dy/dx
sin y	chain rule	cos y · dy/dx
e^y	chain rule	e^y · dy/dx
ln y (y > 0)	chain rule	(1/y) · dy/dx
xy	product rule, then chain rule on y	y + x · dy/dx
y/x	quotient rule	(x · dy/dx − y)/x²
a constant, such as 169	constant rule	0

The pattern is simple: differentiate as if y were x, then attach dy/dx. For x terms there is nothing to attach, because dx/dx = 1.

Back to the circle

Differentiate both sides of x² + y² = 169 with respect to x:

2x + 2y · dy/dx = 0, so dy/dx = −x/y.

At (5, 12): dy/dx = −5/12, matching the explicit answer. At (5, −12): dy/dx = −5/(−12) = +5/12. One formula, both halves of the circle.

The method in five steps

1. Differentiate both sides of the equation with respect to x, term by term.
2. Attach dy/dx to the derivative of every term containing y. Use the product or quotient rule where x and y are multiplied or divided.
3. Collect all terms containing dy/dx on one side and everything else on the other.
4. Factor out dy/dx.
5. Divide to make dy/dx the subject.

To find a slope at a point, check the point is on the curve, then substitute both coordinates.

Why the slope depends on y: a picture

Differentiating x² + 4y² = 25 gives 2x + 8y · dy/dx = 0, so dy/dx = −x/(4y). At P (3, 2): −3/8. At Q (3, −2): +3/8. The y in the formula is exactly what tells the two points apart.

Horizontal and vertical tangent lines

Implicit derivatives usually come out as a fraction, dy/dx = N/D, where N and D contain x and y.

• Horizontal tangent: dy/dx = 0. Set N = 0 (with D ≠ 0), and solve together with the curve’s equation.
• Vertical tangent: the slope is undefined because the tangent line is vertical. Set D = 0 (with N ≠ 0), and solve together with the curve’s equation.
• If N and D are both 0 at a point, the formula gives 0/0 and does not decide the slope there. This needs further analysis.

A solution of N = 0 that is not on the curve is not an answer. Always substitute back into the original equation.

Worked example 1: slope and tangent line at a point

Question. The curve x²y + y³ = 10 passes through (1, 2). Find dy/dx, the slope at (1, 2), and the tangent line there.

1. Check the point. 1² × 2 + 2³ = 2 + 8 = 10. ✓
2. Differentiate both sides. x²y is a product, so use the product rule; y³ needs the chain rule; 10 is a constant. 2x · y + x² · dy/dx + 3y² · dy/dx = 0
3. Collect. Keep the dy/dx terms on the left: x² · dy/dx + 3y² · dy/dx = −2xy.
4. Factor. dy/dx · (x² + 3y²) = −2xy.
5. Divide. dy/dx = −2xy/(x² + 3y²).
6. Substitute both coordinates. At (1, 2): dy/dx = −2(1)(2)/(1 + 3 × 4) = −4/13.
7. Tangent line: y = 2 − (4/13)(x − 1).

Check. Solving the equation numerically for y at x = 1.001 gives y ≈ 1.999692, so the slope is about (1.999692 − 2)/0.001 ≈ −0.308. And −4/13 ≈ −0.3077. They agree.

Common slip. Writing d/dx [x²y] = 2x · dy/dx (differentiating both factors at once) or 2xy (forgetting the second term). The product rule gives two terms.

Worked example 2: horizontal and vertical tangents

Question. For the curve x² − 2xy + 3y² = 18, find dy/dx. Then find every point where the tangent line is horizontal and every point where it is vertical.

1. Differentiate. The middle term −2xy is a product: d/dx [−2xy] = −2y − 2x · dy/dx. 2x − 2y − 2x · dy/dx + 6y · dy/dx = 0
2. Collect and factor. dy/dx · (6y − 2x) = 2y − 2x.
3. Divide and simplify (divide top and bottom by −2): dy/dx = (x − y)/(x − 3y)
4. Horizontal tangents: N = 0, so y = x. Substitute into the curve: x² − 2x² + 3x² = 2x² = 18, so x = ±3. Points: (3, 3) and (−3, −3). Check D: x − 3y = −6 and 6, both nonzero. ✓
5. Vertical tangents: D = 0, so x = 3y. Substitute: 9y² − 6y² + 3y² = 6y² = 18, so y = ±√3. Points: (3√3, √3) and (−3√3, −√3). Check N: x − y = ±2√3, nonzero. ✓

Interpretation. The curve is a tilted ellipse. Its highest point is (3, 3) and its lowest is (−3, −3). Its furthest-right point is (3√3, √3) ≈ (5.20, 1.73).

Worked example 3: an exponential term

Question. The curve e^y + xy = e passes through (0, 1). Find the slope of the curve there and the tangent line.

1. Check the point. e¹ + 0 × 1 = e. ✓
2. Differentiate. e^y needs the chain rule; xy needs the product rule; e is a constant, so its derivative is 0. e^y · dy/dx + y + x · dy/dx = 0
3. Collect and factor. dy/dx · (e^y + x) = −y.
4. Divide. dy/dx = −y/(e^y + x).
5. Substitute (0, 1). dy/dx = −1/(e + 0) = −1/e.
6. Tangent line: y = 1 − x/e.

Common slip. Treating the e on the right as a variable and writing its derivative as e. It is a number, about 2.718, so its derivative is 0.

Common misconceptions

• Forgetting dy/dx on a y term. d/dx [y²] = 2y · dy/dx, not 2y. Without the factor, dy/dx disappears and the algebra cannot be finished.
• Skipping the product rule on xy. d/dx [xy] = y + x · dy/dx. Writing just dy/dx, or just y, loses a term.
• Differentiating a constant to something nonzero. The right side 10, 25 or e becomes 0.
• Substituting only x. dy/dx usually contains y as well. Use both coordinates of a point that is on the curve.
• Leaving dy/dx on both sides. Collect every dy/dx term on one side and factor before dividing.
• Dividing by zero without noticing. If the denominator of dy/dx is 0 at a point, the tangent may be vertical there. Check the numerator too.
• Thinking implicit differentiation needs a function. The curve need not pass the vertical line test. The method gives the slope of the tangent at each point where one exists.
• Accepting a point that is not on the curve. Solving N = 0 alone can give points off the curve. Substitute back.

Where this leads

The same chain-rule move appears again and again. Topic 3.3, differentiating inverse functions, rewrites y = f⁻¹(x) as f(y) = x and differentiates implicitly. Topic 3.6 covers second and higher derivatives; when dy/dx contains y, you find the next derivative by differentiating implicitly again. In Unit 4, related rates differentiate implicitly with respect to time t instead of x. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.