AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/3/3-2)
Unit 3 · Topic 3.2
3.2 Implicit Differentiation
Some curves, like circles, can't be written as a single y = f(x). Implicit differentiation lets you find dy/dx anyway: differentiate both sides with respect to x, treating y as a function of x, so every y term picks up a dy/dx by the chain rule. Then solve for dy/dx.
Key terms
- implicit differentiation
- implicitly defined function
- dy/dx
- chain rule
Why it works
An equation like x² + y² = 25 defines y implicitly: near most points, the curve looks like the graph of some function y(x), even if you can't write it as one formula. Since y is a function of x, the chain rule says d/dx (y²) = 2y·dy/dx.
That's the whole idea. Differentiate x terms as usual. Differentiate y terms as usual, then multiply by dy/dx.
The steps
Follow the same routine every time:
- Differentiate both sides of the equation with respect to x.
- Every time you differentiate an expression in y, multiply by dy/dx.
- Use the product rule on terms like xy: d/dx (xy) = 1·y + x·dy/dx.
- Move all dy/dx terms to one side and everything else to the other.
- Factor out dy/dx and divide.
Derivatives of y terms
These come up again and again. Each is just the usual rule followed by a factor of dy/dx:
| Term | Derivative with respect to x |
|---|---|
| y | dy/dx |
| y³ | 3y²·dy/dx |
| xy | y + x·dy/dx |
| sin y | cos y·dy/dx |
| eʸ | eʸ·dy/dx |
| ln y | (1/y)·dy/dx |
Implicit vs. explicit
For x² + y² = 25 you could solve for y = ±√(25 − x²) and differentiate each half. Implicit differentiation is usually faster, and its answer, −x/y, works on the top and bottom halves at once.
For many curves, solving for y isn't possible at all. Take sin(xy) = y. Differentiating gives cos(xy)·(y + x·dy/dx) = dy/dx, using the chain rule on sine and the product rule on xy. Expanding and collecting: dy/dx·(1 − x cos(xy)) = y cos(xy), so dy/dx = y cos(xy)/(1 − x cos(xy)). You'd never get there by solving for y first.
What the answer looks like
dy/dx usually contains both x and y. To find the slope at a point, plug in both coordinates. For x² + y² = 25, dy/dx = −x/y. At (3, 4) the slope is −3/4; at (3, −4) it is 3/4. Same x, different points, different slopes, which is why you need both coordinates.
Check that the point is actually on the curve before using it. If it doesn't satisfy the original equation, something is wrong with the problem setup or your reading of it.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Slope on a circle
For x² + y² = 25, find dy/dx and the slope of the tangent line at (3, 4).
Show the solutionHide the solution
- Step 1: Differentiate both sides: 2x + 2y·dy/dx = 0.
- Step 2: Solve: dy/dx = −2x/(2y) = −x/y.
- Step 3: At (3, 4): dy/dx = −3/4.
Answer: dy/dx = −x/y; the slope at (3, 4) is −3/4.
- Example 2
Product rule inside implicit differentiation
The curve x³ + xy + y² = 7 passes through (1, 2). Find dy/dx and the equation of the tangent line at (1, 2).
Show the solutionHide the solution
- Step 1: Check the point: 1 + 2 + 4 = 7. Yes, it's on the curve.
- Step 2: Differentiate: 3x² + (y + x·dy/dx) + 2y·dy/dx = 0. The xy term needed the product rule.
- Step 3: Collect dy/dx terms: (x + 2y)·dy/dx = −(3x² + y).
- Step 4: So dy/dx = −(3x² + y)/(x + 2y).
- Step 5: At (1, 2): dy/dx = −(3 + 2)/(1 + 4) = −1.
- Step 6: Tangent line: y − 2 = −1(x − 1).
Answer: dy/dx = −(3x² + y)/(x + 2y); tangent line y − 2 = −(x − 1), or y = −x + 3.
- Example 3
Trap: forgetting dy/dx on a y term
The curve eʸ + y = x + 1 passes through (0, 0). A student differentiates it as eʸ + dy/dx = 1. Find the correct slope at (0, 0).
Show the solutionHide the solution
- Step 1: eʸ is a function of y, and y depends on x, so its derivative is eʸ·dy/dx, not eʸ.
- Step 2: Correct derivative: eʸ·dy/dx + dy/dx = 1.
- Step 3: Factor: dy/dx·(eʸ + 1) = 1, so dy/dx = 1/(eʸ + 1).
- Step 4: At (0, 0): 1/(1 + 1) = 1/2. The student's version would give dy/dx = 1 − e⁰ = 0, which is wrong.
Answer: dy/dx = 1/(eʸ + 1), so the slope at (0, 0) is 1/2.
Common mistakes
- Forgetting dy/dx on y terms, such as writing d/dx (y²) = 2y.
- Treating xy as if only one factor changes. It needs the product rule: y + x·dy/dx.
- Forgetting to differentiate the constant on the right side (it becomes 0), or differentiating it to something else.
On the exam
- Free-response questions often give an implicit curve and dy/dx already computed (“Show that dy/dx = …”), then ask for tangent lines, horizontal or vertical tangents, or the second derivative. If asked to show it, every step must be visible.
- In multiple choice, plugging the point into a correct dy/dx is usually faster than solving for y first.
Connected topics
Videos
Check yourself
4 questions on 3.2 Implicit Differentiation. Pick an answer to see if you got it, and why.
If sin y + y = x², what is dy/dx in terms of x and y?
A curve in the xy-plane is defined by the equation x² + xy + y² = 7. The point (1, 2) lies on the curve.
Described function
Which of the following is an expression for dy/dx?
What is the slope of the line tangent to the curve at the point (1, 2)?
At which of the following points on the curve is the line tangent to the curve horizontal?
0 of 4 answered