Differentiation and Its Applications

Idea of a derivative

Differentiation finds the rate of change of a quantity, or the gradient of a curve at a point. If y = f(x), the derivative is written dy/dx or f'(x).

From first principles: f'(x) = lim as h tends to 0 of [f(x + h) - f(x)] / h.

Rules

1. Constant: d/dx (c) = 0
2. Power rule: d/dx (a x^n) = a n x^(n-1)
3. Sum/difference: differentiate term by term.

Worked examples:

• y = 5x^4 gives dy/dx = 20x^3.
• y = 3x^2 - 7x + 4 gives dy/dx = 6x - 7.
• y = 1/x^2 = x^(-2) gives dy/dx = -2x^(-3) = -2/x^3.
• y = sqrt(x) = x^(1/2) gives dy/dx = 1 / (2 sqrt(x)).

Gradient and tangent

At x = a, gradient = f'(a). The tangent line is y - y1 = m(x - x1).

Example: Find the tangent to y = x^2 + 3x at x = 1. y = 4, so the point is (1, 4). dy/dx = 2x + 3 = 5. Tangent: y - 4 = 5(x - 1), so y = 5x - 1. The normal has gradient -1/5.

Stationary points

Where dy/dx = 0.

1. Solve dy/dx = 0 for x.
2. Find y.
3. Use the second derivative:
   • d2y/dx2 < 0: maximum
   • d2y/dx2 > 0: minimum
   • = 0: test further (possible inflection).

Example: y = x^3 - 3x. dy/dx = 3x^2 - 3 = 0, so x = 1 or -1. d2y/dx2 = 6x. At x = 1: 6 > 0, minimum, y = -2. At x = -1: -6 < 0, maximum, y = 2.

Rates of change

Velocity v = ds/dt and acceleration a = dv/dt = d2s/dt2.

Example: A matatu moves with s = 2t^3 - 9t^2 + 12t metres. v = 6t^2 - 18t + 12. At t = 3, v = 54 - 54 + 12 = 12 m/s. Acceleration = 12t - 18; at t = 3 it is 18 m/s^2. The matatu is momentarily at rest when 6t^2 - 18t + 12 = 0, giving t = 1 or 2.

Optimisation

Used in business and farming, e.g. maximise profit, minimise cost or maximise area.

Example: Profit P = -x^2 + 60x - 400, where x is units sold. dP/dx = -2x + 60 = 0, so x = 30. d2P/dx2 = -2 < 0, so P is a maximum. P = -900 + 1800 - 400 = KES 500 (in thousands if units are given in thousands).

Curve sketching

Find intercepts, stationary points, their nature and behaviour as x gets large. Plot and join smoothly.

Common errors

• Forgetting to reduce the power by 1.
• Differentiating constants incorrectly (they become 0).
• Failing to test the nature of stationary points.
• Missing units in applied answers.

Try it

• Differentiate y = 4x^3 - 2x^2 + x - 9.
• Find the stationary points of y = x^2 - 6x + 5 and state their nature.
• A box with a square base has total surface area 600 cm^2 (open top). Set up an expression for its volume in one variable.