Applications of Differentiation

Apply differentiation at A-Level: tangents, normals, stationary points, optimisation, and rates of change.

Differentiation has powerful applications: finding tangent/normal equations, locating turning points, classifying them, and solving optimisation problems.

Tangents and Normals

At point (a,f(a))(a, f(a)):

  • Tangent: y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x - a)
  • Normal: y−f(a)=−1f′(a)(x−a)y - f(a) = -\frac{1}{f'(a)}(x - a)

Stationary Points

Set f′(x)=0f'(x) = 0 to find stationary points.

Classification

Second derivative test:

  • f′′(x)>0f''(x) > 0: minimum
  • f′′(x)<0f''(x) < 0: maximum
  • f′′(x)=0f''(x) = 0: use sign change of f′(x)f'(x)

Increasing/Decreasing Functions

f′(x)>0f'(x) > 0: increasing. f′(x)<0f'(x) < 0: decreasing.

Optimisation

  1. Form an expression for the quantity to maximise/minimise.
  2. Differentiate and set =0= 0.
  3. Solve and verify (max or min).

Connected Rates of Change

dVdt=dVdr×drdt\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} (chain rule with time).

Worked Example

f(x)=x3−3x2+4f(x) = x^3 - 3x^2 + 4. f′(x)=3x2−6x=3x(x−2)f'(x) = 3x^2 - 6x = 3x(x-2).

Stationary at x=0x = 0 and x=2x = 2.

f′′(0)=−6<0f''(0) = -6 < 0 → max at (0,4)(0, 4). f′′(2)=6>0f''(2) = 6 > 0 → min at (2,0)(2, 0).

Practice Problems

    1. Find the tangent to y=x3y = x^3 at x=2x = 2.
    1. Find and classify stationary points of y=3x4−4x3y = 3x^4 - 4x^3.
    1. A rectangle has perimeter 20. Find dimensions for maximum area.

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Key Takeaways

  • ✓

    Stationary points: f′(x)=0f'(x) = 0.

  • ✓

    Second derivative classifies max/min.

  • ✓

    Optimisation: set up → differentiate → solve → verify.

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