Unit 9: Parametric Equations, Polar Coordinates & Vector-Valued Functions

Derivatives of Parametric Functions

Finding dy/dx and second derivatives of parametrically defined curves.

Advanced25 min lesson2 min readUpdated August 12, 2026Author not yet attributed

Differentiating When Both Coordinates Depend on a Third Variable

For a curve given by x(t) and y(t), neither coordinate is directly a function of the other. Before reading on: since y and x are both functions of the same variable t, could the ordinary chain rule still connect their rates of change to each other?

Definition — Parametric Derivatives

Since dy/dt = (dy/dx)(dx/dt) by the chain rule (treating y as a function of x, and x as a function of t), solving for dy/dx gives dy/dx = (dy/dt)/(dx/dt), valid where dx/dt ≠ 0. The second derivative, d²y/dx², is the derivative of (dy/dx) with respect to x — but dy/dx is itself a function of t, so the chain rule applies again: d²y/dx² = [d/dt(dy/dx)] / (dx/dt).

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Worked Example — Finding a Parametric First Derivative

For x(t) = t²−1, y(t) = t³−3t, find dy/dx at t = 2. dx/dt = 2t, dy/dt = 3t²−3. dy/dx = (3t²−3)/(2t). At t = 2: (12−3)/4 = 9/4.

Worked Example — Finding a Parametric Second Derivative

Continuing the same curve, find d²y/dx² at t = 2. First differentiate dy/dx = (3t²−3)/(2t) with respect to t, using the quotient rule (u=3t²−3, v=2t): d/dt[(3t²−3)/(2t)] = [6t(2t)−(3t²−3)(2)]/(2t)² = [12t²−6t²+6]/(4t²) = (6t²+6)/(4t²). Divide by dx/dt = 2t: d²y/dx² = (6t²+6)/(8t³). At t = 2: (24+6)/64 = 30/64 = 15/32.

Tip

d²y/dx² is never just the second derivative of y divided by the second derivative of x — it requires differentiating the first parametric derivative, dy/dx, with respect to t, then dividing by dx/dt again.

Common Mistakes

Computing d²y/dx² as (d²y/dt²)/(d²x/dt²).
The correct process differentiates dy/dx (not y itself a second time) with respect to t, then divides by dx/dt — two separate applications of the parametric-derivative idea, not a direct ratio of second derivatives.

Key Takeaways

dy/dx for a parametric curve is (dy/dt)/(dx/dt), from the ordinary chain rule.
d²y/dx² requires differentiating dy/dx (as a function of t) and dividing by dx/dt again — not a direct second-derivative ratio.

Summary

The next lesson treats a parametric curve's position as a single vector quantity, connecting these same derivatives to velocity and acceleration.

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