Most tested MM6.3

Applications of Differentiation

This topic covers using differentiation to analyse the properties of a function's graph, such as its slope, turning points, and intervals where it rises or falls. These are fundamental calculus tools for understanding and sketching curves.

Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.

Key points

  • The first derivative, f'(x) or dy/dx, represents the gradient of the tangent to the curve y = f(x) at any given point x.
  • Stationary points (local maxima or minima) are found where the gradient is zero, i.e., by solving the equation f'(x) = 0.
  • The second derivative, f''(x), helps classify stationary points: if f''(a) < 0, it's a local maximum; if f''(a) > 0, it's a local minimum.
  • A function is strictly increasing over an interval where its gradient is consistently positive (f'(x) > 0) and strictly decreasing where its gradient is consistently negative (f'(x) < 0).
  • The gradient of the normal to a curve at a point is the negative reciprocal of the tangent's gradient at that same point, mnormal = -1 / mtangent.
  • A point of inflexion is where a curve's concavity changes. For a simple polynomial like y = x3, a horizontal point of inflexion occurs where f'(x) = 0 and f''(x) = 0.

Diagram

GraphGraph with axes x and y. local maxlocal minxy
A cubic curve has a local maximum and a local minimum (its turning points). Between them the function decreases; outside them it increases.

Formulae

mtangent = f'(x1)

To find the gradient of the tangent to the curve y = f(x) at the point x = x1.

f'(x) = 0

To find the x-coordinates of all stationary points on a curve.

mnormal = -1 / f'(x1)

To find the gradient of the normal to the curve y = f(x) at the point x = x1, provided f'(x1) is not zero.

y - y1 = m(x - x1)

To find the equation of a straight line (tangent or normal) given its gradient 'm' and a point (x1, y1) it passes through.

Definitions

Stationary Point
A point on a curve where the derivative is zero, meaning the tangent is horizontal. It can be a local maximum, local minimum, or a point of inflexion.
Tangent
A straight line that touches a curve at a single point and has the same instantaneous gradient as the curve at that point.
Normal
A straight line that is perpendicular to the tangent at the point where the tangent touches the curve.
Strictly Increasing Function
A function whose value always increases as the input x increases. Its gradient is always positive, f'(x) > 0.

Worked example

Consider the curve with equation y = x3 - 6x2 + 5. Find the equation of the normal to the curve at the point where x = 1.

  1. 1

    First, find the y-coordinate of the point.

    Substitute x=1 into the equation:
    y = (1)3 - 6(1)2 + 5 = 1 - 6 + 5 = 0

    The point is (1, 0).

  2. 2

    Next, find the gradient function by differentiating:

    dy/dx = 3x2 - 12x
  3. 3

    Calculate the gradient of the tangent at x=1:

    mtan = 3(1)2 - 12(1) = 3 - 12 = -9
  4. 4

    The gradient of the normal is the negative reciprocal of the tangent's gradient:

    mnorm = -1 / (-9) = 1/9
  5. 5

    Use the line equation formula y - y1 = m(x - x1) with the point (1, 0) and mnorm = 1/9.

  6. 6
    y - 0 = (1/9)(x - 1)
  7. 7

    Multiply by 9 to clear the fraction:

    9y = x - 1

    This can be rearranged to x - 9y - 1 = 0.

Answer: 9y = x - 1 (or any equivalent form like y = (1/9)x - 1/9)

Common mistakes

  • ×Making sign errors when differentiating or substituting negative numbers. For example, incorrectly calculating y for x=-2 in y = x3 - x as (-2)3 - (-2) = -8 - 2 = -10 instead of -8 + 2 = -6.
  • ×Confusing the normal's gradient with the tangent's. Always remember to find the negative reciprocal (-1/m), not just the reciprocal (1/m) or the negative (-m).
  • ×Arithmetic mistakes when solving the polynomial from f'(x) = 0. Careless factoring or application of the quadratic formula leads to incorrect stationary point locations.

No-calculator tips

  • ✓When solving f'(x) = 0 for a polynomial, always look to factor out a common numerical factor first. Solving 12x2 - 24x + 12 = 0 is easier if you first divide by 12 to get x2 - 2x + 1 = 0.
  • ✓To find the equation of a line with a fractional gradient like m=2/5 passing through (a, b), write y - b = (2/5)(x - a) and immediately multiply the whole equation by 5 to get 5(y - b) = 2(x - a). This clears the fraction and prevents calculation errors.
  • ✓When substituting a value into a polynomial to find a coordinate, calculate the powers first, then multiply, then add/subtract. Following this order reduces the chance of error, e.g., for 2x3 at x=-3, do (-3)3 = -27 first, then 2*(-27) = -54.

Read this topic in the official UAT-UK ESAT guide →

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