Most tested Papers 1 & 2

Differentiation: The Power Rule, Tangents, Normals and Stationary Points

Differentiation finds the gradient function f'(x) = dy/dx, giving the slope of the tangent to a curve and the instantaneous rate of change. This no-calculator TMUA topic tests the power rule for rational n, tangents and normals, and locating and classifying maxima and minima with the second derivative.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • The derivative f'(x) = dy/dx is the gradient of the tangent to the curve at that point, and equals the instantaneous rate of change of y with respect to x.
  • Power rule: multiply by the old power, then reduce the power by 1 (d/dx(xn) = n xn-1); this is valid for any rational n, including negative and fractional powers.
  • Differentiate a sum or difference term by term; a constant multiple stays out front (d/dx(k xn) = k n xn-1) and any constant term differentiates to 0.
  • Always simplify before differentiating: expand brackets, split fractions over their denominator, and rewrite roots and reciprocals as powers (√(x) = x1/2, 1/x2 = x-2).
  • The tangent at (a, f(a)) has gradient f'(a); the normal is perpendicular to it with gradient -1/f'(a).
    product of the two gradients = -1
  • Stationary points occur where f'(x) = 0 (a horizontal tangent); solve that equation to find their x-coordinates.
  • Classify with the second derivative: f''(x) > 0 gives a local minimum and f''(x) < 0 gives a local maximum (TMUA excludes points of inflexion).
  • The second derivative f''(x) = d2y/dx2 is the rate of change of the gradient, found by differentiating f'(x) a second time.

Diagram

xy-2-1012-202y = x^3 - 3xmax (-1, 2)min (1, -2)
Stationary points occur where dy/dx = 0 and the tangent is horizontal: a maximum then a minimum on y = x3 - 3x.

Formulae

d/dx(xn) = n xn-1

Differentiating any power term; works for negative and fractional n once roots and reciprocals are rewritten as powers.

d/dx(k) = 0

Any constant term vanishes when differentiated.

d/dx(a f(x) ± b g(x)) = a f'(x) ± b g'(x)

Differentiate sums and differences term by term, keeping constant multiples out front.

tangent: y - y1 = f'(x1) (x - x1)

Equation of the tangent at (x1, y1), where the gradient is f'(x1).

normal gradient = -1 / f'(x1)

Line perpendicular to the curve at (x1, y1); use it with y - y1 = m(x - x1).

stationary: f'(x) = 0, then test f''(x)

Find stationary points, then apply f''(x) > 0 → minimum, f''(x) < 0 → maximum.

Definitions

Derivative (gradient function)
f'(x) = dy/dx: a new function whose value at any x is the gradient of the tangent to y = f(x) there, equivalently the instantaneous rate of change of y with respect to x.
Stationary point
A point on the curve where f'(x) = 0, so the tangent is horizontal; for TMUA this is a local maximum or a local minimum.
Second derivative
f''(x) = d2y/dx2, the derivative of the derivative; it measures how the gradient is changing and its sign classifies a stationary point.
positive = minimum, negative = maximum
Normal
The straight line through a point on the curve that is perpendicular to the tangent there; its gradient is -1/f'(x1).

Worked examples

1

Find the coordinates of the stationary points of the curve y = 2x3 - 3x2 - 12x + 5 and determine whether each is a maximum or a minimum.

  1. 1

    Differentiate:

    dy/dx = 6x2 - 6x - 12
  2. 2

    Factorise:

    dy/dx = 6(x - 2)(x + 1)
  3. 3
    Solve dy/dx = 0:
    x = 2 or x = -1
  4. 4

    Second derivative:

    d2y/dx2 = 12x - 6
  5. 5
    At x = 2:
    d2y/dx2 = 18 > 0 (minimum)
  6. 6
    y at x = 2:
    y = -15, so (2, -15)
  7. 7
    At x = -1:
    d2y/dx2 = -18 < 0 (maximum)
  8. 8
    y at x = -1:
    y = 12, so (-1, 12)

Answer: Minimum at (2, -15) and maximum at (-1, 12).

2

The curve C has equation y = (x + 3)/√(x) for x > 0. Find the equation of the tangent to C at the point where x = 1.

  1. 1

    Split into powers:

    y = x1/2 + 3x-1/2
  2. 2

    Differentiate:

    dy/dx = (1/2)x-1/2 - (3/2)x-3/2
  3. 3

    Gradient at x = 1:

    dy/dx = 1/2 - 3/2 = -1
  4. 4

    Point on C:

    y = (1 + 3)/1 = 4, so (1, 4)
  5. 5

    Tangent:

    y - 4 = -1(x - 1)
  6. 6

    Simplify:

    y = -x + 5

Answer: y = -x + 5 (equivalently x + y = 5).

Common mistakes

  • ×Getting the power-rule order wrong: you multiply by the old power first, then subtract 1 from it - not the other way round.
  • ×Mishandling negative and fractional powers: differentiating x-2 gives -2x-3 (not -2x-1), and since 1/2 - 1 = -1/2, x1/2 differentiates to (1/2)x-1/2.
  • ×Trying to differentiate a product or quotient term by term without simplifying: you must expand brackets or split the fraction first, as there is no product/quotient shortcut at this level.
  • ×Swapping the tangent and normal gradients: the normal gradient is -1/(tangent gradient), not the tangent gradient itself.
  • ×Reading off max/min from the wrong sign: f''(x) > 0 is a minimum (curve bends up, a 'smile') and f''(x) < 0 is a maximum (a 'frown').

No-calculator tips

  • ✓Keep every answer exact - leave surds and fractions as they are (e.g. gradient = -3/2), since TMUA options are stated in exact form.
  • ✓Factorise f'(x) before solving f'(x) = 0: for example 6x2 - 6x - 12 = 6(x - 2)(x + 1) reads off the roots x = 2 and x = -1 instantly, no quadratic formula needed.
  • ✓To classify a stationary point you often need only the sign of f''(x), so compute the y-coordinate only when the question actually asks for coordinates.
  • ✓Use elimination: substitute a candidate x back into f'(x) to check it gives 0, or test whether a suggested tangent passes through the given point, to reject wrong options fast.
  • ✓Translate 'rate of change' straight into dy/dx evaluated at the stated point - the wording is just asking you to differentiate and substitute.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.The curve y = x2 - 6x + 5 passes through the point where x = 4. What is the gradient of the NORMAL to the curve at that point?

  • A. 2
  • B. 1/2
  • C. -1/2
  • D. -2
Show answer

Answer: C — -1/2

y' = 2x - 6, so at x = 4 the tangent gradient is 2; the normal gradient is the negative reciprocal, -1/2. Choosing 2 gives the tangent instead of the normal, and -2 negates without taking the reciprocal.

Q2.The curve y = x3 - 3x2 - 9x + 5 has stationary points. Using the second derivative, which statement is correct?

  • A. Local maximum at x = -1 and local minimum at x = 3
  • B. Local minimum at x = -1 and local maximum at x = 3
  • C. Stationary points at x = 1 and x = -3
  • D. Local minimum at x = 3 only
  • E. The curve has no stationary points
Show answer

Answer: A — Local maximum at x = -1 and local minimum at x = 3

y' = 3x2 - 6x - 9 = 3(x + 1)(x - 3), so x = -1 or x = 3; since y'' = 6x - 6 is negative at x = -1 (max) and positive at x = 3 (min). Option 3 comes from mis-factoring the sign, and option 2 confuses the second-derivative sign test.

Q3.For the curve y = x + 4/x (x ≠ 0), use the second derivative to classify its stationary points. Which statement is correct?

  • A. Local minimum at (2, 4) only; there are no other stationary points
  • B. Local maximum at (2, 4) and local minimum at (-2, -4)
  • C. Stationary points at (2, 4) and (-2, -4), both local minima
  • D. Local minimum at (2, 4) and local maximum at (-2, -4)
Show answer

Answer: D — Local minimum at (2, 4) and local maximum at (-2, -4)

y' = 1 - 4/x2 = 0 gives x2 = 4, so x = ±2 (both roots); y'' = 8/x3 is positive at x = 2 (minimum at (2, 4)) and negative at x = -2 (maximum at (-2, -4)). Option 1 keeps only the positive root, the classic error here.

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