Implicit differentiation – tangents to curves F(x, y) = 0

What it shows

A curve such as x² + xy + y² = 3 is not the graph of a single function, yet near most of its points it defines y as a function of x. Differentiating both sides with respect to x, using the chain rule for terms in y and the product rule for xy, gives dy/dx; in general dy/dx = −Fₓ/F_y. The tangent is horizontal where Fₓ = 0 and vertical where F_y = 0, and where both vanish the point is singular. The curve is drawn numerically on a grid, so the values are close approximations. A parametrization x(t), y(t) traces the same curve and gives the same slope y′(t)/x′(t).

How to use

Choose a Curve and set k with the slider, or pick Type your own F(x, y) = 0 and enter an equation. Drag the red point P along the curve to read dy/dx, the tangent line and d²y/dx²; P snaps to the green and purple points where the tangent is horizontal or vertical. Tick Show ∇F = (Fₓ, F_y) to see why the tangent is perpendicular to the gradient. Press Trace with parametrization to compare y′(t)/x′(t) with −Fₓ/F_y.

Parameters you can change

• Curve Circle x² + y² = k², Ellipse x²/k² + y²/4 = 1, Tilted ellipse x² + xy + y² = k, Hyperbola x² − y² = k, Folium of Descartes x³ + y³ = 3kxy, Cubic curve y² = x³ − 3x + k, Type your own F(x, y) = 0
• Parameter k 0.5–4
• Typed equation F(x, y) = 0
• Show points with horizontal and vertical tangents

Questions to explore

1. On x² + xy + y² = 3, at which points is the tangent horizontal, and how can you find them algebraically?
2. Why does implicit differentiation give no single tangent at the origin of the folium x³ + y³ = 9xy?
3. For the circle x² + y² = 9, why is dy/dx = −x/y undefined at (3, 0)?