AP Calculus BC · Unit 3: Composite, Implicit & Inverse Derivatives
3.6 Calculating Higher-Order Derivatives
What are higher-order derivatives?
Differentiating a function multiple times gives higher-order derivatives. The first derivative f'(x) gives slope/rate of change; the second derivative f''(x) is the derivative of f'(x), showing concavity and acceleration. Notation includes f''(x), y'', d²y/dx², or f⁽ⁿ⁾(x) for the nth derivative. Just keep differentiating the previous result.
Implicit differentiation for higher orders
To find d²y/dx² implicitly, differentiate the equation once to get dy/dx, then differentiate again with respect to x, substituting the expression for dy/dx you already found. This often introduces both x, y, and dy/dx terms — simplify by replacing dy/dx with its known expression before solving for d²y/dx².
Higher-order derivatives of parametric/vector functions
For parametric curves, d²y/dx² is NOT simply the derivative of dy/dx with respect to t. Instead, compute dy/dx = (dy/dt)/(dx/dt), then d²y/dx² = [d/dt(dy/dx)] / (dx/dt). Always divide by dx/dt again — a common error is skipping this final division.
Key terms
- Second derivative notation
- f''(x), y'', or d²y/dx² — the derivative of the first derivative.
- nth derivative notation
- f⁽ⁿ⁾(x) or dⁿy/dxⁿ, found by differentiating n times.
- Implicit second derivative method
- Differentiate the equation twice with respect to x, substituting the known dy/dx expression before solving for d²y/dx².
- Parametric second derivative formula
- d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt) — divide by dx/dt twice, not d²x/dt².
- Physical meaning of second derivative
- If position is s(t), s'(t) is velocity and s''(t) is acceleration.
Try an AP-style question
For f(x) = x⁴ − 3x², what is f''(x)?
- 12x² − 6x
- 4x³ − 6x
- 24x
- 12x² − 6
Show the answer
D. f' = 4x³ − 6x; f'' = 12x² − 6.
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