Form 5 · Chapter 2

Application of Differentiation

Apply differentiation to tangents, normals, rates of change and maximum/minimum problems.

Tangents and normals

The gradient of the tangent to a curve at a point equals the value of dy/dx there. The normal is perpendicular to the tangent, so its gradient is −1/(dy/dx). Once you have a gradient m and a point (x1, y1), the line is y − y1 = m(x − x1).

Key formula

tangent gradient = dy/dx · normal gradient = −1/(dy/dx) · rate of change: dy/dt = (dy/dx)(dx/dt) · small change: δy ≈ (dy/dx) δx

Rates of change and small changes

When two quantities vary with time, the chain rule links their rates: dy/dt = (dy/dx)(dx/dt). For a small change in x, the approximate change in y is δy ≈ (dy/dx) δx.

Worked example

For y = x2 − 3x + 2 at x = 2: dy/dx = 2x − 3 = 1, so the tangent gradient is 1 and the normal gradient is −1. For maximum and minimum, y = x2 − 6x + 5 gives dy/dx = 2x − 6 = 0 at x = 3, where y = 9 − 18 + 5 = −4 (a minimum since d2y/dx2 = 2 > 0).

Optimisation problems

To find a maximum or minimum, set dy/dx = 0 to locate the stationary point, then use the second derivative to confirm its nature. This is the standard method for problems on largest area, least cost, and similar situations.

Remember

  • Normal gradient = −1 divided by the tangent gradient.
  • Connected rates: dy/dt = (dy/dx)(dx/dt).
  • For max/min, solve dy/dx = 0 then test with d2y/dx2.

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