AP Calc Implicit Differentiation Explained Simply

Implicit differentiation is how you find dy/dx when y is not isolated, as in x² + y² = 25. You differentiate both sides with respect to x, treat y as a function of x (so every y term picks up a dy/dx from the chain rule), then solve for dy/dx. That is the whole idea; the rest is careful algebra.

Implicit differentiation is tested in both AP Calculus AB and BC, and it is the foundation for related rates and for derivatives of inverse functions.

When do you use implicit differentiation?

Use it whenever an equation mixes x and y and solving for y would be messy or impossible.

Even when you could solve for y, as with a circle, implicit differentiation is usually faster and avoids the ± square root.

Why does y get a dy/dx?

The key idea is that y is secretly a function of x. When you differentiate y², you are differentiating [y(x)]², so the chain rule gives 2y · (dy/dx).

Compare these side by side:

• d/dx of x² = 2x
• d/dx of y² = 2y(dy/dx)
• d/dx of sin(y) = cos(y)(dy/dx)
• d/dx of xy = x(dy/dx) + y (product rule)

If you remember one thing, remember this: every time you differentiate a y term, multiply by dy/dx.

How to do implicit differentiation step by step

1. Differentiate both sides with respect to x. Do every term, including constants (which become 0).
2. Attach dy/dx to every derivative of a y term. Use the product rule whenever x and y are multiplied.
3. Move all dy/dx terms to one side. Move everything else to the other side.
4. Factor out dy/dx.
5. Divide to solve for dy/dx.
6. Substitute a point if the question asks for a slope. Use both the x- and y-coordinates.

Worked example 1: a circle

Find dy/dx for x² + y² = 25, then find the slope of the tangent line at (3, 4).

1. Differentiate: 2x + 2y(dy/dx) = 0.
2. Isolate: 2y(dy/dx) = −2x.
3. Solve: dy/dx = −x/y.
4. At (3, 4): dy/dx = −3/4.

The tangent line at (3, 4) is y − 4 = −(3/4)(x − 3).

Notice that the derivative depends on both x and y. That is normal for implicit derivatives, and it is why you need the full point to get a slope.

Worked example 2: a product term

Find dy/dx for x³ + xy + y² = 7, and evaluate it at (1, 2).

First, confirm the point is on the curve: 1 + 2 + 4 = 7. It is.

1. Differentiate each term:
   • x³ becomes 3x²
   • xy becomes x(dy/dx) + y by the product rule
   • y² becomes 2y(dy/dx)
   • 7 becomes 0
2. Write it out: 3x² + x(dy/dx) + y + 2y(dy/dx) = 0.
3. Collect dy/dx terms: (x + 2y)(dy/dx) = −3x² − y.
4. Solve: dy/dx = −(3x² + y)/(x + 2y).
5. At (1, 2): dy/dx = −(3 + 2)/(1 + 4) = −1.

The most common error here is forgetting the product rule on xy and writing just x(dy/dx). That single slip changes the answer.

Worked example 3: a trig equation

Find dy/dx for sin(y) = x.

1. Differentiate: cos(y)(dy/dx) = 1.
2. Solve: dy/dx = 1/cos(y).

This is exactly how the derivative of arcsin x is derived. Since y = arcsin x, you can rewrite cos(y) in terms of x to get 1/√(1 − x²). That connection to inverse functions is tested on the AP exam, and our AP Calculus formula sheet lists the inverse trig derivatives you should know.

How do you find the second derivative implicitly?

Some AP questions ask for d²y/dx². Differentiate your first derivative again, remembering that y is still a function of x, then substitute your expression for dy/dx.

For the circle, dy/dx = −x/y. Using the quotient rule:

d²y/dx² = −[y · 1 − x(dy/dx)] / y²

Substitute dy/dx = −x/y:

d²y/dx² = −[y + x²/y] / y² = −(y² + x²)/y³

Because x² + y² = 25 on this curve, the second derivative simplifies to −25/y³.

That last substitution, using the original equation to simplify, is a move AP readers love to see, and it often appears in free-response questions.

Horizontal and vertical tangent lines

Implicit derivatives usually come out as a fraction. That makes tangent-line questions quick.

• Horizontal tangent: dy/dx = 0, so set the numerator equal to zero (with the denominator nonzero).
• Vertical tangent: dy/dx is undefined, so set the denominator equal to zero (with the numerator nonzero).

After you find the condition, substitute it back into the original equation to find the actual points on the curve. Students often stop after finding the condition and lose the final point.

For the circle, dy/dx = −x/y. Horizontal tangents occur where x = 0, at (0, 5) and (0, −5). Vertical tangents occur where y = 0, at (5, 0) and (−5, 0).

Common implicit differentiation mistakes

• Forgetting dy/dx after differentiating a y term.
• Skipping the product rule on terms like xy or x²y.
• Treating a constant on the right side as anything other than 0 after differentiating.
• Plugging in only x when the derivative depends on both x and y.
• Not checking that the given point is actually on the curve.

How implicit differentiation connects to related rates

Related rates is implicit differentiation with respect to time instead of x. Every variable gets a d/dt instead of a dy/dx. Once implicit differentiation feels natural, read our guide to AP Calc related rates and the setup will feel familiar.

For how these skills earn credit on the free response, see our AP Calculus FRQ tips. If you are deciding between courses, AP Calculus AB vs BC compares the two. You can estimate your score from practice tests with the AP Calculus BC score calculator.

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