← Differentiation methods
6 Parametric differentiation
Checked against 2022-2025: a parametric question has appeared in every single paper, always at Merit-Excellence. Treat it as a core topic, not an extra.Parametric differentiation
First derivative \frac{dy}{dx}=\frac{dy/dt}{dx/dt} Differentiate x and y separately with respect to t, then divide. Second derivative \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt} Treat dy/dx as a new function of t: differentiate it again with respect to t, then divide by dx/dt a second time. ✕d²y/dx² is NOT (d²y/dt²) ÷ (d²x/dt²)
The tempting shortcut — differentiate x and y each twice with respect to t, then divide — gives the wrong answer almost every time. The correct method treats dy/dx itself as a function of t, differentiates THAT with respect to t, and only then divides by dx/dt again. Two separate divisions by dx/dt, not one division of two second derivatives. Worked example Gradient at a given value of t A curve has x = t² + 1 and y = 2t³. Find dy/dx, and its value when t = 2. dx/dt = 2t. dy/dt = 6t². dy/dx = (dy/dt) ÷ (dx/dt) = 6t² ÷ 2t = 3t (for t ≠ 0). At t = 2: dy/dx = 3(2) = 6. ✓ dy/dx = 3t, so the gradient at t = 2 is 6. Worked example Points of inflection, parametrically A curve is defined by x = t² and y = t³ − 3t. Find the value(s) of t at any points of inflection. dx/dt = 2t. dy/dt = 3t² − 3, so dy/dx = (3t² − 3) ÷ 2t. Differentiate dy/dx with respect to t (quotient rule, u = 3t² − 3, v = 2t): d/dt(dy/dx) = (6t·2t − (3t² − 3)·2) / (2t)² = (12t² − 6t² + 6) / 4t² = (6t² + 6) / 4t². d²y/dx² = [(6t² + 6)/4t²] ÷ 2t = (6t² + 6) / 8t³ = 3(t² + 1) / 4t³. Setting the numerator to zero: t² + 1 = 0 has no real solution. ✓ No points of inflection exist — d²y/dx² = 3(t² + 1)/(4t³) is never zero, since t² + 1 > 0 for all real t.Method
- Differentiate x and y separately with respect to t.
- dy/dx = (dy/dt) ÷ (dx/dt). Substitute a given value of t directly if only the gradient is wanted.
- For d²y/dx², differentiate dy/dx AGAIN with respect to t, then divide by dx/dt a second time.
- To find where the gradient equals a target value, or where a stationary/inflection point occurs, set the appropriate derivative to that value (or to zero) and solve for t — then substitute back for (x, y) if the coordinates are asked for.