Short on time? This page is the recap. For the figure and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- Differentiate implicitly and write dy/dx = N/D, with N and D in terms of x and y.
- A critical point is a point on the curve where dy/dx = 0 or dy/dx does not exist.
- Setting N = 0 or D = 0 gives a line or curve, not a point. Solve it together with the curve’s equation.
- Check the other part of the fraction at every point you find.
- Use the sign of dy/dx (in terms of x and y) for increasing and decreasing, and the sign of d²y/dx² for concavity and the second derivative test.
Key relationships
| At a point on the curve | Meaning |
|---|---|
| N = 0, D ≠ 0 | dy/dx = 0: horizontal tangent; candidate for a relative max or min of y |
| D = 0, N ≠ 0 | dy/dx undefined: vertical tangent; a critical point, but not a max or min of y |
| N = 0, D = 0 | 0/0: the formula decides nothing; investigate another way |
| dy/dx = 0 and d²y/dx² < 0 | relative maximum of y |
| dy/dx = 0 and d²y/dx² > 0 | relative minimum of y |
| dy/dx changes + to − (or − to +) | relative maximum (or minimum) of y |
Shortcut for y″ at a horizontal tangent: every term containing y′ is 0. For y′ = N/D with N = 0 at the point, y″ = N′/D there, where N′ is the derivative of N with respect to x, evaluated with y′ = 0.
Assumptions behind the method
- The point really lies on the curve: substitute it into the original equation.
- Near a point with D ≠ 0, the curve is locally the graph of a function y(x), so the usual tests apply to that piece.
- d²y/dx² may contain x, y and dy/dx. You need both coordinates (and dy/dx) to evaluate it.
Mistakes to avoid
- Stopping at “y = 2x” instead of finding points on the curve.
- Calling 0/0 a horizontal tangent.
- Forgetting that dy/dx not existing also makes a critical point.
- Treating y as a constant when differentiating again.
- Evaluating with x alone when one x gives two points.
- “It has a max”: say “y has a relative maximum of … at x = …”.
Quick self-check
- dy/dx = (x − 1)/(y + 3). Where could the tangent be vertical? (Where y = −3 on the curve, provided x ≠ 1 there)
- For y² = x² − 4x + 13, differentiating 2y·y′ = 2x − 4 again gives 2(y′)² + 2y·y″ = 2. Classify the horizontal tangent at (2, 3). (y′ = 0 there, so 6y″ = 2 and y″ = 1/3 > 0: relative minimum of y)
- dy/dx = x/y. On which quadrants is y decreasing? (II and IV, where x and y have opposite signs)
Next: practice questions.