Σ TMUA Practice
Home›Topics›Differentiation

Differentiation

Derivatives of x^n, gradients, tangents and normals, stationary points, increasing and decreasing functions, second derivatives.

Spec MM6.1-MM6.3, MM8.5Section 1 Part 1: AS-level pure mathematics22 questions26 real past questions
📖 Revision notes🧩 Solution bank⚡ Quick-fire this topicFormula cardsSet 1 questions (PDF)Set 1 solutionsSet 2 questions (PDF)Set 2 solutions

Click an option to answer. A correct first attempt marks the question secure; a wrong one marks it needs work and opens the worked solution. Keys A-H answer the question nearest the top of the screen.

1Differentiation · spec MM6.2 · Paper 1 style · ★☆☆

Find \(\dfrac{dy}{dx}\) when \(y = \dfrac{(3x - 2)^2}{x^2}\).

  1. A\(\dfrac{12}{x^2} + \dfrac{8}{x^3}\)
  2. B\(\dfrac{6}{x} - \dfrac{12}{x^2}\)
  3. C\(-\dfrac{12}{x^2} + \dfrac{8}{x^3}\)
  4. D\(\dfrac{12}{x^2} - \dfrac{8}{x^3}\)
  5. E\(\dfrac{6(3x-2)}{x^2}\)
  6. F\(9 - \dfrac{12}{x}\)
Worked solution

\(y = 9 - 12x^{-1} + 4x^{-2}\), so \(y' = 12x^{-2} - 8x^{-3}\).

Full method — how and why
The spec only allows differentiating powers of \(x\), so the expression must first be written as a sum of such powers - expand and divide term by term.
  1. Expand the numerator: \((3x-2)^2 = 9x^2 - 12x + 4\).
    There is no quotient or chain rule available, so the fraction must be dismantled.
  2. Divide each term by \(x^2\): \(y = 9 - 12x^{-1} + 4x^{-2}\).
    \(\frac{9x^2}{x^2} = 9\), \(\frac{-12x}{x^2} = -12x^{-1}\), \(\frac{4}{x^2} = 4x^{-2}\).
  3. Differentiate term by term with \(\frac{d}{dx}x^n = nx^{n-1}\): \(0 + 12x^{-2} - 8x^{-3}\).
    \(-12 \times (-1) = +12\) and \(4 \times (-2) = -8\); the constant \(9\) differentiates to \(0\).
  4. Write in fraction form: \(\dfrac{12}{x^2} - \dfrac{8}{x^3}\).
    Match the form used in the options.
Where it goes wrong
  • Differentiating the bracket directly as \(\frac{6(3x-2)}{x^2}\) - that would need the quotient rule (and is wrong anyway).
  • Sign slips: differentiating \(x^{-1}\) gives \(-x^{-2}\), so \(-12x^{-1}\) gives \(+12x^{-2}\).
Takeaway Rewrite as \(\sum ax^n\) before differentiating; negative powers change sign when differentiated.
Answer D · Differentiation (MM6.2) · question DIF-01
2Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

Find the coordinates of the stationary points of \(y = x^3 - 3x^2 - 9x + 5\) and state their nature.

  1. Amaximum at \((1, -6)\), minimum at \((-3, -22)\)
  2. Bminimum at \((-1, 10)\), maximum at \((3, -22)\)
  3. Cmaximum at \((-1, 10)\), minimum at \((3, -22)\)
  4. Dmaximum at \((-1, 0)\), minimum at \((3, -32)\)
  5. Emaximum at \((-1, 10)\) only
Worked solution

\(y' = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)\). \(y'' = 6x - 6\): negative at \(x = -1\) (max, \(y = -1 - 3 + 9 + 5 = 10\)), positive at \(x = 3\) (min, \(y = 27 - 27 - 27 + 5 = -22\)).

Full method — how and why
Stationary points from \(f' = 0\), their nature from the sign of \(f''\), and the \(y\)-values by substitution.
  1. Differentiate: \(y' = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)\).
    Factor out the \(3\) first - it makes the factorisation obvious.
  2. Solve \(y' = 0\): \(x = 3\) and \(x = -1\).
    Stationary points are where the gradient vanishes.
  3. Second derivative: \(y'' = 6x - 6\). At \(x = -1\), \(y'' = -12 < 0\) (maximum); at \(x = 3\), \(y'' = 12 > 0\) (minimum).
    Negative second derivative means the curve bends downwards - a maximum.
  4. Find the \(y\)-values: \(y(-1) = -1 - 3 + 9 + 5 = 10\); \(y(3) = 27 - 27 - 27 + 5 = -22\).
    Substitute into the original function, not the derivative.
Where it goes wrong
  • Swapping the natures - for a positive cubic the maximum always comes first (smaller \(x\)).
  • Substituting into \(y'\) instead of \(y\) for the coordinates.
Takeaway \(f' = 0\) locates stationary points; the sign of \(f''\) classifies them; substitute into \(f\) for the heights.
Answer C · Differentiation (MM6.3) · question DIF-02
3Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The tangent to \(y = x^2 - 4x + 1\) at the point where \(x = 3\) meets the \(y\)-axis at

  1. A\((0, -2)\)
  2. B\((0, 8)\)
  3. C\((0, 4)\)
  4. D\((0, 1)\)
  5. E\((0, -5)\)
  6. F\((0, -8)\)
Worked solution

\(y' = 2x - 4 = 2\) at \(x = 3\); the point is \((3, -2)\). Tangent: \(y + 2 = 2(x - 3)\), so at \(x = 0\), \(y = -8\).

Full method — how and why
Tangent line: gradient from the derivative, point from the curve, then find where it meets the axis.
  1. Differentiate: \(y' = 2x - 4\); at \(x = 3\) the gradient is \(2\).
    The derivative evaluated at a point gives the tangent's gradient there.
  2. Find the point: \(y(3) = 9 - 12 + 1 = -2\), so the tangent touches at \((3, -2)\).
    The tangent passes through the curve at that \(x\) - use the original function.
  3. Write the tangent: \(y + 2 = 2(x - 3)\).
    Point-gradient form with the point and gradient just found.
  4. Set \(x = 0\): \(y + 2 = -6\), so \(y = -8\). The tangent meets the \(y\)-axis at \((0, -8)\).
    The \(y\)-axis is \(x = 0\).
Where it goes wrong
  • Using \(y(3) = -2\) as the intercept and answering \((0,-2)\).
  • Sign slip: \(y = 2x - 8\), not \(2x + 8\).
Takeaway Tangent \(=\) gradient from \(f'\) \(+\) point from \(f\); then substitute \(x = 0\) for the intercept.
Answer F · Differentiation (MM6.3) · question DIF-03
4Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

Find the equation of the normal to the curve \(y = \sqrt{x}\) at the point \((4, 2)\).

  1. A\(y = -4x + 16\)
  2. B\(y = -4x + 18\)
  3. C\(y = -4x + 2\)
  4. D\(y = \dfrac14 x + 1\)
  5. E\(y = -\dfrac14 x + 3\)
  6. F\(y = 4x - 14\)
Worked solution

\(y' = \frac{1}{2\sqrt x} = \frac14\) at \(x = 4\), so the normal has gradient \(-4\): \(y - 2 = -4(x - 4)\), i.e. \(y = -4x + 18\).

Full method — how and why
Normal line: the negative reciprocal of the tangent's gradient, through the same point.
  1. Write \(y = x^{1/2}\) and differentiate: \(y' = \frac12 x^{-1/2} = \dfrac{1}{2\sqrt x}\).
    Convert the root to a fractional power before differentiating.
  2. At \(x = 4\): \(y' = \dfrac{1}{2 \times 2} = \dfrac14\).
    \(\sqrt4 = 2\).
  3. Normal gradient \(= -\dfrac{1}{1/4} = -4\).
    Normal is perpendicular to the tangent: negative reciprocal.
  4. Line through \((4,2)\): \(y - 2 = -4(x - 4) \Rightarrow y = -4x + 18\).
    \(-4 \times -4 = +16\), plus the \(2\).
Where it goes wrong
  • Using the tangent gradient \(\frac14\) and choosing \(y = \frac14x + 1\).
  • Arithmetic on the constant: \(y = -4x + 16\) is offered.
Takeaway Normal gradient \(= -1/f'(x_1)\); then substitute the point into the point-gradient form.
Answer B · Differentiation (MM6.3) · question DIF-04
5Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

For what range of values of \(x\) is \(f(x) = 2x^3 - 15x^2 + 24x + 1\) decreasing?

  1. A\(-4 < x < -1\)
  2. B\(2 < x < 3\)
  3. C\(x > 1\)
  4. D\(x < 1\) or \(x > 4\)
  5. E\(x < 4\)
  6. F\(1 < x < 4\)
Worked solution

\(f'(x) = 6x^2 - 30x + 24 = 6(x - 1)(x - 4) < 0\) for \(1 < x < 4\).

Full method — how and why
'Decreasing' is the condition \(f'(x) < 0\) - a quadratic inequality once the derivative is found.
  1. Differentiate: \(f'(x) = 6x^2 - 30x + 24\).
    Term by term.
  2. Factorise: \(6(x^2 - 5x + 4) = 6(x-1)(x-4)\).
    Taking out the \(6\) makes the roots visible.
  3. Solve \(f'(x) < 0\): the upward parabola is negative between its roots, so \(1 < x < 4\).
    The gradient is negative exactly where the function falls.
Where it goes wrong
  • Solving \(f'(x) > 0\) and answering \(x < 1\) or \(x > 4\) - which is offered.
  • Forgetting the factor \(6\) when factorising, giving the wrong roots.
Takeaway Decreasing \(\Leftrightarrow f' < 0\); find the roots of \(f'\) and use the shape of its graph.
Answer F · Differentiation (MM6.3) · question DIF-05
6Differentiation · spec MM6.1 · Paper 1 style · ★☆☆

The displacement \(s\) metres of a particle at time \(t\) seconds is \(s = t^3 - 6t^2 + 9t\). At what times is the particle momentarily at rest?

  1. A\(t = 2\) only
  2. B\(t = 1\) and \(t = 3\)
  3. C\(t = 1\) only
  4. D\(t = 3\) only
  5. E\(t = 0\) and \(t = 3\)
  6. F\(t = 0\), \(t = 1\) and \(t = 3\)
Worked solution

Velocity \(= \frac{ds}{dt} = 3t^2 - 12t + 9 = 3(t - 1)(t - 3) = 0\) at \(t = 1\) and \(t = 3\).

Full method — how and why
'At rest' means velocity zero, and velocity is the derivative of displacement.
  1. Differentiate the displacement: \(v = \dfrac{ds}{dt} = 3t^2 - 12t + 9\).
    Velocity is the rate of change of displacement.
  2. Factorise: \(3(t^2 - 4t + 3) = 3(t-1)(t-3)\).
    Take the \(3\) out first.
  3. Set \(v = 0\): \(t = 1\) and \(t = 3\).
    'Momentarily at rest' is exactly \(v = 0\).
Where it goes wrong
  • Setting \(s = 0\) instead of \(v = 0\), which gives \(t = 0\) and \(t = 3\) - an offered option.
  • Forgetting one of the two roots.
Takeaway Displacement \(\to\) velocity \(\to\) acceleration by differentiating; 'at rest' means the velocity is zero, not the displacement.
Answer B · Differentiation (MM6.1) · question DIF-06
7Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The curve \(y = ax^2 + bx\) has a stationary point at \((2, -8)\). Find \(a + b\).

  1. A\(-6\)
  2. B\(-2\)
  3. C\(10\)
  4. D\(2\)
  5. E\(-10\)
  6. F\(6\)
Worked solution

\(y' = 2ax + b = 0\) at \(x = 2\): \(b = -4a\). \(y(2) = 4a + 2b = 4a - 8a = -4a = -8 \Rightarrow a = 2\), \(b = -8\). \(a + b = -6\).

Full method — how and why
A stationary point gives two conditions: the derivative vanishes there, and the point is on the curve.
  1. Differentiate: \(y' = 2ax + b\). At \(x = 2\): \(4a + b = 0\), so \(b = -4a\).
    Stationary means the gradient is zero at that \(x\).
  2. The point is on the curve: \(y(2) = 4a + 2b = -8\).
    Substitute \(x = 2\) into the original function and set it equal to the given \(y\).
  3. Substitute \(b = -4a\): \(4a - 8a = -4a = -8\), so \(a = 2\) and \(b = -8\).
    One equation in one unknown after substitution.
  4. Compute \(a + b = 2 + (-8) = -6\).
    The question asks for the sum.
Where it goes wrong
  • Using only one of the two conditions.
  • Sign error giving \(a = -2\) and the answer \(6\).
Takeaway 'Stationary point at \((p,q)\)' is two equations: \(f'(p) = 0\) and \(f(p) = q\).
Answer A · Differentiation (MM6.3) · question DIF-07
8Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

A rectangle has perimeter \(40\). What is its maximum possible area?

  1. A\(80\)
  2. B\(50\)
  3. C\(160\)
  4. D\(400\)
  5. E\(100\)
  6. F\(200\)
Worked solution

Sides \(x\) and \(20 - x\): \(A = 20x - x^2\), \(A' = 20 - 2x = 0\) at \(x = 10\), giving \(A = 100\) (a square).

Full method — how and why
A standard optimisation: express the area in one variable using the constraint, then differentiate.
  1. Let one side be \(x\); the perimeter condition \(2x + 2y = 40\) gives \(y = 20 - x\).
    Use the constraint to eliminate the second variable - optimisation needs a single variable.
  2. Write the area: \(A = x(20 - x) = 20x - x^2\).
    A quadratic in \(x\).
  3. Differentiate and solve: \(A' = 20 - 2x = 0\) at \(x = 10\).
    Or complete the square: \(A = 100 - (x-10)^2\), which shows the maximum immediately.
  4. Compute the area: \(A = 10 \times 10 = 100\) (a square).
    \(A'' = -2 < 0\) confirms a maximum.
Where it goes wrong
  • Using \(x + y = 40\) instead of \(2x + 2y = 40\), giving \(400\) - an offered option.
  • Reporting the side length \(10\) rather than the area.
Takeaway For fixed perimeter, the rectangle of greatest area is the square - but show it via the constraint and the derivative.
Answer E · Differentiation (MM6.3) · question DIF-09
9Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The curve \(y = x^3 + px^2 + qx\) has stationary points at \(x = 1\) and \(x = 3\). Find \(q\).

  1. A\(3\)
  2. B\(12\)
  3. C\(9\)
  4. D\(-9\)
  5. E\(-6\)
  6. F\(6\)
Worked solution

\(y' = 3x^2 + 2px + q = 3(x - 1)(x - 3) = 3x^2 - 12x + 9\). So \(q = 9\) (and \(p = -6\)).

Full method — how and why
Known stationary points mean known roots of \(f'\). Compare the factorised derivative with its expanded form.
  1. Differentiate: \(y' = 3x^2 + 2px + q\).
    A cubic's derivative is a quadratic.
  2. The stationary points are its roots, so \(y' = 3(x-1)(x-3)\).
    The leading coefficient must be \(3\) to match \(3x^2\).
  3. Expand: \(3(x^2 - 4x + 3) = 3x^2 - 12x + 9\).
    Multiply out to compare coefficients.
  4. Compare constants: \(q = 9\) (and comparing \(x\) terms gives \(2p = -12\), so \(p = -6\)).
    The question asks only for \(q\).
Where it goes wrong
  • Forgetting the leading \(3\) and getting \(q = 3\).
  • Answering \(-6\), which is \(p\).
Takeaway Stationary points are roots of \(f'\): build \(f'\) in factorised form with the correct leading coefficient.
Answer C · Differentiation (MM6.3) · question DIF-10
10Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The line \(y = 4x + c\) is a tangent to \(y = x^2 + 2x + 5\). Find \(c\).

  1. A\(4\)
  2. B\(-4\)
  3. C\(1\)
  4. D\(5\)
  5. E\(6\)
  6. F\(3\)
Worked solution

\(2x + 2 = 4 \Rightarrow x = 1\), \(y = 8\). \(8 = 4 + c \Rightarrow c = 4\).

Full method — how and why
Tangency via calculus: the gradients must match, which locates the point of contact.
  1. Differentiate the curve: \(y' = 2x + 2\).
    The tangent's gradient equals the curve's gradient at the contact point.
  2. Set it equal to the line's gradient: \(2x + 2 = 4\), so \(x = 1\).
    This finds where the tangent touches.
  3. Find the \(y\)-value on the curve: \(y(1) = 1 + 2 + 5 = 8\).
    The contact point is on the curve.
  4. The point must also be on the line: \(8 = 4(1) + c\), so \(c = 4\).
    Substituting the contact point into the line's equation gives \(c\).
Where it goes wrong
  • Using the discriminant route but mis-expanding and getting \(c = 1\) or \(5\) (both offered).
  • Substituting into the line the \(y\)-value from the wrong \(x\).
Takeaway Tangency: equal gradients give the \(x\); equal \(y\)-values give the constant.
Answer A · Differentiation (MM6.3) · question DIF-12
11Differentiation · spec MM6.2 · Paper 1 style · ★☆☆

Find \(\dfrac{d^2y}{dx^2}\) when \(y = \dfrac{1}{x^2} + 4\sqrt{x}\).

  1. A\(\dfrac{3}{x^4} - x^{-\frac32}\)
  2. B\(\dfrac{6}{x^4} + x^{-\frac32}\)
  3. C\(\dfrac{6}{x^4} - 2x^{-\frac32}\)
  4. D\(\dfrac{6}{x^4} - x^{-\frac32}\)
  5. E\(-\dfrac{6}{x^4} - x^{-\frac32}\)
  6. F\(\dfrac{2}{x^3} + 2x^{-\frac12}\)
Worked solution

\(y' = -2x^{-3} + 2x^{-1/2}\); \(y'' = 6x^{-4} - x^{-3/2}\).

Full method — how and why
Two differentiations of powers, with negative and fractional indices. Rewrite first.
  1. Rewrite: \(y = x^{-2} + 4x^{1/2}\).
    Roots and reciprocals must become powers before the rule applies.
  2. First derivative: \(y' = -2x^{-3} + 4 \times \frac12 x^{-1/2} = -2x^{-3} + 2x^{-1/2}\).
    \(-2 \times x^{-2-1}\) and \(\frac12 \times 4 = 2\).
  3. Second derivative: \(y'' = -2(-3)x^{-4} + 2\left(-\frac12\right)x^{-3/2} = 6x^{-4} - x^{-3/2}\).
    Differentiate again, watching the signs: \(-3\) times \(-2\) gives \(+6\).
  4. Write it as \(\dfrac{6}{x^4} - x^{-3/2}\).
    Matching the option format.
Where it goes wrong
  • Losing a sign and answering \(\frac{6}{x^4} + x^{-3/2}\).
  • Stopping after the first derivative.
Takeaway Differentiate twice, carefully: each negative index gains one more negative factor.
Answer D · Differentiation (MM6.2) · question DIF-14
12Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The curve \(y = x^2 + kx\) (with \(k < 0\)) crosses the \(x\)-axis at the origin and at one other point \(P\). The gradient of the curve at \(P\) is \(6\). Find \(k\).

  1. A\(6\)
  2. B\(-6\)
  3. C\(-12\)
  4. D\(-3\)
  5. E\(3\)
  6. F\(-2\)
Worked solution

\(P = (-k, 0)\). \(y' = 2x + k = -2k + k = -k = 6\), so \(k = -6\).

Full method — how and why
Find the second crossing point in terms of \(k\), then impose the gradient condition there.
  1. Find the \(x\)-intercepts: \(x^2 + kx = x(x + k) = 0\), so \(x = 0\) or \(x = -k\). Thus \(P = (-k, 0)\).
    With \(k < 0\), \(-k\) is positive - the curve crosses to the right of the origin.
  2. Differentiate: \(y' = 2x + k\).
    The gradient at any point.
  3. Evaluate at \(P\): \(y'(-k) = -2k + k = -k\).
    Substituting \(x = -k\).
  4. Set equal to \(6\): \(-k = 6\), so \(k = -6\).
    Consistent with the given condition \(k < 0\) ✓.
Where it goes wrong
  • Using \(x = k\) for the second intercept, giving \(k = 2\) or \(3k = 6\).
  • Answering \(6\) and ignoring the stated sign of \(k\).
Takeaway Find the point in terms of the parameter first, then apply the gradient condition; check any sign restriction at the end.
Answer B · Differentiation (MM6.3) · question DIF-17
13Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

Find the minimum value of \(x + \dfrac{4}{x}\) for \(x > 0\).

  1. A\(5\)
  2. B\(2\sqrt2\)
  3. C\(4\)
  4. D\(0\)
  5. E\(2\)
  6. F\(8\)
Worked solution

Derivative \(1 - \frac{4}{x^2} = 0\) at \(x = 2\); value \(2 + 2 = 4\) (second derivative \(\frac{8}{x^3} > 0\), so a minimum).

Full method — how and why
Minimising \(x + \frac4x\). Differentiate, solve, and confirm it is a minimum.
  1. Write as powers: \(y = x + 4x^{-1}\) and differentiate: \(y' = 1 - 4x^{-2}\).
    \(\frac{d}{dx}(4x^{-1}) = -4x^{-2}\).
  2. Solve \(y' = 0\): \(1 = \dfrac{4}{x^2} \Rightarrow x^2 = 4 \Rightarrow x = 2\) (taking the positive root since \(x > 0\)).
    The restriction \(x > 0\) discards \(x = -2\).
  3. Confirm a minimum: \(y'' = 8x^{-3} > 0\) for \(x > 0\).
    Positive second derivative means the curve bends upwards.
  4. Evaluate: \(y = 2 + \dfrac42 = 4\).
    (Also follows from AM-GM: \(x + \frac4x \geq 2\sqrt{4} = 4\).)
Where it goes wrong
  • Giving the minimising \(x = 2\) instead of the minimum value \(4\) - both are offered.
  • Including \(x = -2\), where the expression equals \(-4\) but \(x > 0\) is required.
Takeaway \(x + \frac kx \geq 2\sqrt k\) for \(x > 0\), with equality at \(x = \sqrt k\) - worth recognising instantly.
Answer C · Differentiation (MM6.3) · question DIF-18
14Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

Find the sum of the \(y\)-coordinates of the stationary points of the curve \(y = 2x^3 - 9x^2 + 12x - 3\).

  1. A\(2\)
  2. B\(0\)
  3. C\(5\)
  4. D\(4\)
  5. E\(3\)
  6. F\(1\)
Worked solution

\(y' = 6x^2 - 18x + 12 = 6(x - 1)(x - 2)\): stationary points at \(x = 1\) (\(y = 2 - 9 + 12 - 3 = 2\)) and \(x = 2\) (\(y = 16 - 36 + 24 - 3 = 1\)). Sum \(3\).

Full method — how and why
Find both stationary points and add their \(y\)-values - two substitutions after one factorisation.
  1. Differentiate: \(y' = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)\).
    Take out the \(6\), then factorise.
  2. Stationary points at \(x = 1\) and \(x = 2\).
    Roots of the derivative.
  3. Compute the \(y\)-values: \(y(1) = 2 - 9 + 12 - 3 = 2\); \(y(2) = 16 - 36 + 24 - 3 = 1\).
    Substitute into the original cubic, carefully: \(2(8) = 16\) and \(9(4) = 36\).
  4. Add: \(2 + 1 = 3\).
    The question asks for the sum of the \(y\)-coordinates.
Where it goes wrong
  • Adding the \(x\)-coordinates (\(1 + 2 = 3\)) - the same answer here by coincidence, but the wrong method.
  • Arithmetic slips in \(y(2)\); recompute \(16 - 36 = -20\), \(-20 + 24 = 4\), \(4 - 3 = 1\).
Takeaway Two stationary points mean two substitutions into the original function - the derivative gives only the positions.
Answer E · Differentiation (MM6.3) · question DIF-20
15Differentiation · spec MM6.2 · Paper 1 style · ★★☆

Given \(f(x) = x^{\frac{3}{2}} - 6x^{\frac{1}{2}}\) for \(x > 0\), find the value of \(x\) at which \(f\) has its minimum.

  1. A\(2\)
  2. B\(4\)
  3. C\(1\)
  4. D\(6\)
  5. E\(\sqrt{2}\)
  6. F\(3\)
Worked solution

\(f'(x) = \frac32 x^{1/2} - 3x^{-1/2} = \frac{3}{2\sqrt x}(x - 2) = 0\) at \(x = 2\); \(f'\) changes from negative to positive there, so it is a minimum.

Full method — how and why
Differentiating fractional powers. Factorising the derivative makes the sign change easy to see.
  1. Differentiate: \(f'(x) = \frac32 x^{1/2} - 3 \times \frac12 x^{-1/2} = \frac32 x^{1/2} - 3x^{-1/2}\).
    \(\frac{d}{dx}x^{3/2} = \frac32 x^{1/2}\) and \(\frac{d}{dx}(6x^{1/2}) = 3x^{-1/2}\).
  2. Factor out \(\dfrac{3}{2\sqrt x}\): \(f'(x) = \dfrac{3}{2\sqrt x}\left(x - 2\right)\).
    \(\frac32 x^{1/2} = \frac{3}{2\sqrt x}\cdot x\) and \(3x^{-1/2} = \frac{3}{2\sqrt x}\cdot 2\). Factorising isolates the sign-changing part.
  3. Set \(f' = 0\): since \(\frac{3}{2\sqrt x} > 0\) for \(x > 0\), the only root is \(x = 2\).
    A positive factor can never make the product zero.
  4. Check it is a minimum: \(f'\) is negative for \(x < 2\) and positive for \(x > 2\), so the curve falls then rises.
    The sign of \(f'\) is the sign of \((x-2)\) - immediate from the factorised form.
Where it goes wrong
  • Making index errors: \(\frac{d}{dx}x^{1/2} = \frac12 x^{-1/2}\), not \(\frac12 x^{1/2}\).
  • Squaring both sides carelessly when solving \(\frac32\sqrt x = \frac{3}{\sqrt x}\) and getting \(x = 4\).
Takeaway Factor a common power out of the derivative: the remaining bracket controls the sign.
Answer A · Differentiation (MM6.2) · question DIF-08
16Differentiation · spec MM6.3 · Paper 2 style · ★★☆

Which of the following statements about \(f(x) = x^3 - 3x\) is/are true?

  1. I\(f\) is increasing for all \(x > 1\).
  2. II\(f\) has a local maximum at \(x = -1\).
  3. III\(f(x) \geq -2\) for all real \(x\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

\(f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\): positive for \(x > 1\) (I true); \(f'' = 6x < 0\) at \(x = -1\) (II true). III false: the local minimum \(f(1) = -2\) is not a global minimum since \(f(x) \to -\infty\) as \(x \to -\infty\) (e.g. \(f(-3) = -18\)).

Full method — how and why
Three claims about a cubic. The derivative settles the first two; the third is about global behaviour, where cubics are unbounded.
  1. Differentiate: \(f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\).
    The roots \(\pm1\) are the stationary points.
  2. I: for \(x > 1\) both factors are positive, so \(f' > 0\) and \(f\) is increasing. True.
    Sign analysis of the factorised derivative.
  3. II: \(f'' = 6x\), so \(f''(-1) = -6 < 0\): local maximum at \(x = -1\). True.
    Negative second derivative means a local maximum.
  4. III: the local minimum is \(f(1) = -2\), but as \(x \to -\infty\), \(f(x) \to -\infty\) - for instance \(f(-3) = -27 + 9 = -18\). False.
    A local minimum is not a global one; odd-degree polynomials are unbounded below.
  5. I and II only.
    Match to the options.
Where it goes wrong
  • Assuming the local minimum value bounds the whole function.
  • Confusing the maximum and minimum positions - for \(+x^3\) the maximum is at the smaller \(x\).
Takeaway Cubics have no global maximum or minimum; 'local' claims need \(f'\), 'global' claims need the end behaviour.
Answer E · Differentiation (MM6.3) · question DIF-11
17Differentiation · spec MM6.1 · Paper 1 style · ★★☆

\(f(x)\) is a polynomial with \(f'(x) = (x - 2)^2(x + 1)\). Which of the following is true?

  1. A\(f\) has local minima at \(x = -1\) and \(x = 2\)
  2. B\(f\) has no turning points
  3. C\(f\) has a local minimum at \(x = -1\) and no turning point at \(x = 2\)
  4. D\(f\) has a local maximum at \(x = -1\) and a local minimum at \(x = 2\)
  5. E\(f\) has a local minimum at \(x = -1\) and a local maximum at \(x = 2\)
Worked solution

\(f'\) changes sign from negative to positive at \(x = -1\) (local minimum). At \(x = 2\) the factor \((x-2)^2\) does not change sign, so \(f'\) stays positive: a stationary point but not a turning point (a point of inflexion).

Full method — how and why
Reading a factorised derivative. A squared factor does not change the sign, so it gives a stationary point that is not a turning point.
  1. The stationary points are the roots of \(f'\): \(x = -1\) and \(x = 2\).
    \(f'(x) = 0\) at both.
  2. Near \(x = -1\): \((x-2)^2 > 0\) always, so the sign of \(f'\) is the sign of \((x+1)\) - negative then positive.
    A sign change from \(-\) to \(+\) means the function falls then rises: a local minimum.
  3. Near \(x = 2\): \((x+1) > 0\) there, and \((x-2)^2 \geq 0\), so \(f' \geq 0\) on both sides - no sign change.
    A repeated (even-power) factor touches zero without crossing, so the gradient does not change sign.
  4. Conclusion: local minimum at \(x = -1\); stationary but not a turning point at \(x = 2\).
    That is a stationary point of inflexion.
Where it goes wrong
  • Assuming every root of \(f'\) is a turning point.
  • Using \(f''(2) = 0\) to conclude 'maximum' or 'minimum' - a zero second derivative is inconclusive.
Takeaway Odd-power factors of \(f'\) change sign (turning points); even-power factors do not (inflexions).
Answer C · Differentiation (MM6.1) · question DIF-13
18Differentiation · spec MM6.3 · Paper 1 style · ★★☆

The volume of a cuboid with a square base of side \(x\) is \(V = 4x^2 - x^3\) for \(0 < x < 4\). Find the maximum volume.

  1. A\(16\)
  2. B\(\dfrac{64}{27}\)
  3. C\(8\)
  4. D\(\dfrac{32}{3}\)
  5. E\(\dfrac{256}{27}\)
  6. F\(\dfrac{128}{27}\)
Worked solution

\(V' = 8x - 3x^2 = x(8 - 3x) = 0 \Rightarrow x = \frac83\). \(V = \frac{64}{9}\cdot 4 - \frac{512}{27} = \frac{768 - 512}{27} = \frac{256}{27}\).

Full method — how and why
Maximise a cubic on an interval. Differentiate, find the interior stationary point, and evaluate.
  1. Differentiate: \(V' = 8x - 3x^2 = x(8 - 3x)\).
    Factorise to find the roots easily.
  2. Roots at \(x = 0\) and \(x = \frac83\). Only \(x = \frac83\) is inside \(0 < x < 4\) with \(V'\) changing from \(+\) to \(-\) there.
    \(x = 0\) is an endpoint (and gives zero volume), so the maximum is at \(\frac83\).
  3. Evaluate: \(V = 4\left(\frac83\right)^2 - \left(\frac83\right)^3 = \frac{256}{9} - \frac{512}{27}\).
    \(\left(\frac83\right)^2 = \frac{64}{9}\), so \(4 \times \frac{64}{9} = \frac{256}{9}\); and \(\left(\frac83\right)^3 = \frac{512}{27}\).
  4. Common denominator \(27\): \(\dfrac{768 - 512}{27} = \dfrac{256}{27}\).
    \(\frac{256}{9} = \frac{768}{27}\).
Where it goes wrong
  • Arithmetic slips with the fractions - \(\frac{128}{27}\) and \(\frac{64}{27}\) are both offered.
  • Taking \(x = 0\) as the maximiser.
Takeaway Optimise on an interval: check interior stationary points and whether they are maxima; keep fractions exact.
Answer E · Differentiation (MM6.3) · question DIF-15
19Differentiation · spec MM6.3 · Paper 1 style · ★★☆

The tangent to \(y = x^3\) at the point \((a, a^3)\), where \(a \neq 0\), meets the curve again at the point with \(x\)-coordinate

  1. A\(a/2\)
  2. B\(-2a\)
  3. C\(2a\)
  4. D\(0\)
  5. E\(-a\)
  6. F\(-a/2\)
Worked solution

Tangent: \(y = 3a^2 x - 2a^3\). Solve \(x^3 - 3a^2 x + 2a^3 = 0\); \(x = a\) is a double root, so \((x - a)^2(x + 2a) = 0\): the other intersection is \(x = -2a\).

Full method — how and why
A tangent meeting the curve again. The tangency shows up as a repeated root, which lets you factorise the cubic without solving it.
  1. Find the tangent at \((a, a^3)\): \(y' = 3x^2\), so the gradient is \(3a^2\) and the line is \(y = 3a^2(x - a) + a^3 = 3a^2x - 2a^3\).
    Point-gradient form, then simplify.
  2. Set the tangent equal to the curve: \(x^3 = 3a^2 x - 2a^3\), i.e. \(x^3 - 3a^2x + 2a^3 = 0\).
    Intersections satisfy both equations.
  3. \(x = a\) is a repeated root (that is what tangency means), so factor out \((x-a)^2\): \(x^3 - 3a^2x + 2a^3 = (x-a)^2(x + 2a)\).
    Check by comparing the constant term: \((-a)^2(2a) = 2a^3\) ✓.
  4. The remaining root is \(x = -2a\).
    That is where the tangent cuts the curve again.
Where it goes wrong
  • Forgetting that tangency gives a double root and trying to factor out only one \((x-a)\).
  • Sign error giving \(x = 2a\).
Takeaway A tangent meets the curve in a repeated root; factor \((x - a)^2\) out and read off the third root.
Answer B · Differentiation (MM6.3) · question DIF-16
20Differentiation · spec MM6.1 · Paper 2 style · ★★☆

Which of the following statements is/are true for every function \(f\) that has a derivative at every real number?

  1. IIf \(f'(a) = 0\) then \(f\) has a maximum or a minimum at \(x = a\).
  2. IIIf \(f\) has a local minimum at \(x = a\) then \(f'(a) = 0\).
  3. IIIIf \(f'(x) > 0\) for all \(x\), then \(f(3) > f(2)\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I is false: \(f(x) = x^3\) has \(f'(0) = 0\) but no turning point. II is true (a stationary point is required at an interior local extremum of a differentiable function). III is true: a strictly increasing function has \(f(3) > f(2)\).

Full method — how and why
Three general claims about differentiable functions. The first is the classic false one; the others are standard facts.
  1. I: test \(f(x) = x^3\) at \(a = 0\): \(f'(0) = 0\) but the function is increasing throughout, so there is no maximum or minimum. False.
    A stationary point can be a point of inflexion.
  2. II: at an interior local minimum of a differentiable function the derivative must vanish (otherwise you could move downhill in one direction). True.
    This is Fermat's stationary-point theorem, and is why we solve \(f' = 0\) in the first place.
  3. III: if \(f' > 0\) everywhere then \(f\) is strictly increasing, so \(f(3) > f(2)\). True.
    Positive gradient throughout means the value rises as \(x\) rises.
  4. II and III only.
    Match to the options.
Where it goes wrong
  • Accepting I because 'we find maxima and minima by setting \(f' = 0\)' - that finds candidates, not guarantees.
  • Doubting II by thinking of endpoints, which the word 'every real number' excludes.
Takeaway \(f'(a) = 0\) is necessary but not sufficient for a turning point; inflexions are the counterexample.
Answer G · Differentiation (MM6.1) · question DIF-19
21Differentiation · spec MM6.3 · Paper 2 style · ★★☆

Let \(f\) be a polynomial. \(P\): '\(f'(x) > 0\) for all real \(x\)'. \(Q\): '\(f(x) = 0\) has exactly one real root'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Worked solution

If \(f' > 0\) everywhere, \(f\) is strictly increasing and (being an odd-degree polynomial) crosses the axis exactly once: \(P\) is sufficient. But \(f(x) = x^3\) has exactly one root and \(f'(0) = 0\), so \(P\) is not necessary.

Full method — how and why
A necessary/sufficient question about polynomials. Test each direction, and look for a boundary case to break the converse.
  1. Sufficiency: if \(f'(x)>0\) for all \(x\), then \(f\) is strictly increasing, so it takes each value at most once - at most one root.
    A strictly increasing function cannot return to a value it has already passed.
  2. Moreover such an \(f\) must have odd degree (an even-degree polynomial turns), so it goes from \(-\infty\) to \(+\infty\) and crosses the axis. Exactly one root: \(P \Rightarrow Q\). Sufficient.
    Both halves are needed: 'at most one' plus 'at least one'.
  3. Necessity: find an \(f\) with exactly one root but \(f'\) not always positive. Take \(f(x) = x^3\): one root at \(0\), but \(f'(0) = 0\).
    A stationary point of inflexion satisfies \(Q\) without satisfying \(P\).
  4. So \(P\) is sufficient but not necessary.
    Match to the list.
Where it goes wrong
  • Claiming \(P\) is necessary because 'one root suggests increasing' - \(x^3\) and also \(x^2\)-type shifts break that.
  • Forgetting that \(f'>0\) forces odd degree, so 'at least one root' still needs justifying.
Takeaway Strictly increasing \(\Rightarrow\) at most one root; but a single root does not force a strictly positive derivative (\(x^3\)).
Answer C · Differentiation (MM6.3) · question X2-02
22Differentiation · spec MM6.3 · Paper 2 style · ★★☆

The function \(f\) has \(f'(x) = (x - 1)(x - 3)^2\). Which of the following statements is/are true?

  1. I\(f\) has a local minimum at \(x = 1\).
  2. II\(f\) has a local maximum at \(x = 3\).
  3. III\(f\) is increasing for all \(x > 1\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

\(f'\) changes from negative to positive at \(x = 1\) (minimum). At \(x = 3\) the squared factor keeps \(f' \geq 0\) on both sides, so no turning point: II is false, and \(f' \geq 0\) for \(x > 1\) (zero only at \(3\)) makes III true.

Full method — how and why
Classify stationary points by whether \(f'\) changes sign - a squared factor does not change sign.
  1. \(f'(x) = (x-1)(x-3)^2\), so \(f'=0\) at \(x=1\) and \(x=3\).
    Stationary points are the roots of \(f'\).
  2. Sign of \(f'\): \((x-3)^2 \geq 0\) always, so the sign is that of \((x-1)\). Negative for \(x<1\), positive for \(x>1\) (zero at \(x=3\)).
    An even-power factor contributes no sign change.
  3. I: \(f'\) goes negative \(\to\) positive at \(x=1\), so a local minimum. True.
    Falling then rising.
  4. II: at \(x=3\), \(f'\) is positive on both sides, so no turning point - a stationary point of inflexion. False.
    The squared factor is precisely why there is no maximum here.
  5. III: for \(x>1\), \(f' \geq 0\) with equality only at the isolated point \(x=3\), so \(f\) is increasing. True.
    An isolated zero of the derivative does not interrupt increase.
  6. So I and III.
    Match to the list.
Where it goes wrong
  • Assuming every root of \(f'\) is a turning point and alternating min/max - the repeated root breaks the alternation.
  • Rejecting III because \(f'(3)=0\); increase needs \(f' \geq 0\) with no interval of equality.
Takeaway Odd-power factors of \(f'\) change sign (turning point); even-power factors do not (stationary inflexion).
Answer F · Differentiation (MM6.3) · question X2-16

How this topic has been examined

Real TMUA questions on Differentiation, newest first. Links open the official paper at the question and the online marking page.

2023 · Paper 1Question 14cubic with three distinct real rootsquestion papersit online
2022 · Paper 1Question 15largest rectangle between two parabolasquestion papersit online
2022 · Paper 2Question 1stationary points of 3x^4 + 4x^3 + 6x^2 - 5question papersit online
2021 · Paper 1Question 11fraction of an interval where f decreasesquestion papersit online
2021 · Paper 2Question 8exists c with p'(c) = 0question papersit online
2020 · Paper 1Question 1derivative of (x^3 - 5x^2) / (2x sqrt x)question papersit online
2019 · Paper 1Question 7gradient at x = -1 as a function of qquestion papersit online
2019 · Paper 1Question 18shortest distance from a parabola to a linequestion papersit online
2019 · Paper 2Question 14statements p(a), p(b), p'(a), p'(b)question papersit online
October 2018 · Paper 1Question 92x^3 - 3x^2 - 12x + c: three real rootsquestion papersit online
October 2018 · Paper 1Question 11normal to y = 10 - x^2question papersit online
October 2018 · Paper 1Question 13graph of f': which point is a local minimum of fquestion papersit online
October 2018 · Paper 2Question 1f(x) = (x^3 - 4x)/... for x > 0question papersit online
October 2018 · Paper 2Question 15x^3 + 3qx^2 + 2 with three distinct rootsquestion papersit online
November 2017 · Paper 1Question 2f''(1) for (2/x - 1/(2x^2))^2question papersit online
November 2017 · Paper 1Question 10greatest gradient of a normal as p variesquestion papersit online
November 2017 · Paper 1Question 16one function increasing, the other decreasingquestion papersit online
November 2017 · Paper 2Question 1derivative of (1 - 3x)^2 / (2x^(3/2))question papersit online
Practice papers · Paper 1Question 3normal to y = 2/x^2: length PQquestion papersit online
Practice papers · Paper 1Question 12largest cylinder inside a spherequestion papersit online
Practice papers · Paper 1Question 15least gradient at x = 1 as a variesquestion papersit online
Specimen papers · Paper 1Question 15smallest possible value as x variesquestion papersit online
Practice papers · Paper 1Question 18where (1 - x)/x^(2/3) decreasesquestion papersit online
Practice papers · Paper 2Question 2derivative of (x^2 + 5)(2x)/x^(3/4)question papersit online
Specimen papers · Paper 2Question 2gradient at x = 2question papersit online
Practice papers · Paper 2Question 6sequence f(n+1) = x f(n)'question papersit online