Short on time? This page is the recap. For explanations, the ellipse diagram and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- An implicit equation mixes x and y, for example x² + 4y² = 25. You do not need to solve for y.
- Treat y as a function of x. Every y term is then a composite, so the chain rule attaches a factor dy/dx.
- Terms that multiply x and y need the product rule; terms that divide them need the quotient rule.
- Method: differentiate both sides → collect dy/dx terms → factor out dy/dx → divide.
- The answer usually contains x and y. To evaluate it, use both coordinates of a point that is on the curve.
- Write dy/dx = N/D. Horizontal tangent: N = 0, D ≠ 0. Vertical tangent: D = 0, N ≠ 0. Both 0: the formula does not decide.
Key relationships
| Term | d/dx of the term |
|---|---|
| yⁿ | n yⁿ⁻¹ · dy/dx |
| sin y | cos y · dy/dx |
| cos y | −sin y · dy/dx |
| e^y | e^y · dy/dx |
| ln y (y > 0) | (1/y) · dy/dx |
| xy | y + x · dy/dx |
| x²y | 2xy + x² · dy/dx |
| a constant | 0 |
Assumptions behind the method
- Near the point, the curve is locally the graph of a differentiable function y(x).
- The point you substitute must satisfy the original equation.
- Dividing by D is only valid where D ≠ 0; where D = 0, investigate separately.
Mistakes to avoid
- Writing d/dx [y²] = 2y, with no dy/dx.
- Writing d/dx [xy] = dy/dx, or just y. The product rule gives both terms.
- Differentiating the constant on the right to anything but 0.
- Substituting only x into a dy/dx that contains y.
- Leaving dy/dx terms on both sides instead of collecting and factoring.
- Using a point from N = 0 that is not on the curve.
Quick self-check
- Find the slope of x² + y² = 10 at (1, 3). (dy/dx = −x/y = −1/3)
- Find dy/dx if y⁴ = x. (1/(4y³))
- Find the slope of xy = 6 at (2, 3). (y + x · dy/dx = 0, so dy/dx = −y/x = −3/2)
- A curve has dy/dx = (x + 1)/(y − 2). Where could the tangent be vertical? (Where y = 2 and x ≠ −1, at points on the curve)
Next: practice questions.