Higher-Order Derivatives

Find second and higher derivatives for AP Calculus AB. Interpret concavity and acceleration.

Higher-order derivatives are obtained by differentiating repeatedly. The second derivative has important applications in concavity and acceleration.

Notation

f′′(x)=d2ydx2f''(x) = \frac{d^2y}{dx^2} (second derivative).

f′′′(x)=d3ydx3f'''(x) = \frac{d^3y}{dx^3} (third derivative).

Interpretation

  • f′(x)f'(x): rate of change / velocity.
  • f′′(x)f''(x): rate of change of the rate / acceleration / concavity.

Concavity

  • f′′(x)>0f''(x) > 0: concave up (cup shape).
  • f′′(x)<0f''(x) < 0: concave down (cap shape).
  • f′′(x)=0f''(x) = 0: possible inflection point.

Inflection Points

Where concavity changes: f′′(x)=0f''(x) = 0 AND f′′f'' changes sign.

Worked Example

f(x)=x4−6x2f(x) = x^4 - 6x^2. f′(x)=4x3−12xf'(x) = 4x^3 - 12x. f′′(x)=12x2−12f''(x) = 12x^2 - 12.

f′′(x)=0f''(x) = 0: 12(x2−1)=012(x^2 - 1) = 0 → x=±1x = \pm 1.

f′′f'' changes sign at both → inflection points at x=−1x = -1 and x=1x = 1.

Practice Problems

    1. Find f′′(x)f''(x) for f(x)=sin⁡(2x)f(x) = \sin(2x).
    1. Find inflection points of f(x)=x3−3x2+4f(x) = x^3 - 3x^2 + 4.

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Key Takeaways

  • ✓

    f′′f'' determines concavity.

  • ✓

    Inflection points where f′′f'' changes sign.

  • ✓

    Second derivative test: min if f′′>0f'' > 0, max if f′′<0f'' < 0.

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