Question Simulator

AS & A Level Further Mathematics (9231)

Differentiation Banker

17 questions · about 5% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 22 · Q15 marksMaclaurin series
  • 2026 May/June · Paper 22 · Q4 (a, b)8 marksImplicit differentiation

2025

  • 2025 Oct/Nov · Paper 22 · Q3 (a, b)9 marksImplicit differentiation
  • 2025 Oct/Nov · Paper 22 · Q7 (a)3 marksDerivatives of hyperbolic & inverse functions
  • 2025 May/June · Paper 22 · Q6 (b, c, d)12 marksDerivatives of hyperbolic & inverse functions · Parametric differentiation

2024

  • 2024 Oct/Nov · Paper 22 · Q2 (a, b)7 marksImplicit differentiation
  • 2024 Oct/Nov · Paper 22 · Q6 (d)3 marksMaclaurin series
  • 2024 Oct/Nov · Paper 22 · Q7 (a)3 marksDerivatives of hyperbolic & inverse functions
  • 2024 May/June · Paper 22 · Q24 marksMaclaurin series
  • 2024 May/June · Paper 22 · Q3 (a, b)7 marksParametric differentiation

2023

  • 2023 Oct/Nov · Paper 22 · Q15 marksMaclaurin series
  • 2023 Oct/Nov · Paper 22 · Q2 (a, b)7 marksParametric differentiation
  • 2023 Oct/Nov · Paper 22 · Q5 (a)4 marksDerivatives of hyperbolic & inverse functions
  • 2023 May/June · Paper 22 · Q8 (b, c)11 marksParametric differentiation

2022

  • 2022 Oct/Nov · Paper 22 · Q2 (a, b)7 marksImplicit differentiation
  • 2022 Oct/Nov · Paper 22 · Q3 (b)2 marksMaclaurin series
  • 2022 May/June · Paper 22 · Q5 (a, b)10 marksImplicit differentiation · Maclaurin series

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Implicit differentiation · Core · Paper 2 style · 5 marks
  1. (a) The curve C has equation x³ + xy + y² = 7. Show that the point (1, 2) lies on C and find dy/dx at this point. [3]
  2. (b) Find the equation of the tangent to C at (1, 2). [2]
Show answer
  1. (a) 1 + 2 + 4 = 7; dy/dx = −1
    B1 verification; M1 3x² + y + x(dy/dx) + 2y(dy/dx) = 0; A1 dy/dx = −(3x² + y)/(x + 2y) = −5/5 = −1.
  2. (b) x + y = 3
    M1 y − 2 = −1(x − 1); A1 x + y = 3.
Q2Implicit differentiation · Core · Paper 2 style · 5 marks
  1. (a) The curve C has equation x²y + y² = 6. Find dy/dx in terms of x and y. [3]
  2. (b) Find the equation of the tangent to C at the point (1, 2), giving your answer in the form ax + by = c. [2]
Show answer
  1. (a) dy/dx = −2xy/(x² + 2y)
    M1 product rule on x²y; M1 2xy + x²(dy/dx) + 2y(dy/dx) = 0; A1 result.
  2. (b) 4x + 5y = 14
    M1 gradient −4/5; A1 4x + 5y = 14.
Q3Implicit differentiation · Core · Paper 2 style · 5 marks
  1. (a) The curve C has equation xeʸ + y = 2. Find the gradient of C at the point where it crosses the x-axis. [3]
  2. (b) Find the equation of the normal to C at this point. [2]
Show answer
  1. (a) −1/3 (at the point (2, 0))
    B1 y = 0 gives x = 2; M1 eʸ + xeʸ(dy/dx) + dy/dx = 0; A1 dy/dx = −eʸ/(xeʸ + 1) = −1/3.
  2. (b) y = 3x − 6
    M1 normal gradient 3; A1 y = 3(x − 2).
Q4Implicit differentiation · Standard · Paper 2 style · 7 marks
  1. (a) The curve C has equation x² + xy + y² = 7. Find the value of dy/dx at the point (1, 2). [3]
  2. (b) Find the value of d²y/dx² at (1, 2). [4]
Show answer
  1. (a) −4/5
    M1 2x + y + x(dy/dx) + 2y(dy/dx) = 0; A1 dy/dx = −(2x + y)/(x + 2y); A1 −4/5.
  2. (b) −42/125
    M1 differentiate the implicit equation again: 2 + 2(dy/dx) + x(d²y/dx²) + 2(dy/dx)² + 2y(d²y/dx²) = 0; M1 substitute x = 1, y = 2, dy/dx = −4/5; A1 5(d²y/dx²) = −(2 − 8/5 + 32/25) = −42/25; A1 d²y/dx² = −42/125.
Q5Implicit differentiation · Standard · Paper 2 style · 7 marks
  1. (a) The curve C has equation e^(xy) = x + y. Show that (0, 1) lies on C and that dy/dx = 0 there. [3]
  2. (b) Find d²y/dx² at (0, 1) and hence determine the nature of this stationary point. [4]
Show answer
  1. (a) e⁰ = 1 = 0 + 1; dy/dx = 0
    B1 verification; M1 e^(xy)(y + x(dy/dx)) = 1 + dy/dx; A1 at (0, 1): 1 = 1 + dy/dx so dy/dx = 0.
  2. (b) d²y/dx² = 1; minimum point
    M1 differentiate again: e^(xy)(y + x y′)² + e^(xy)(2y′ + x y″) = y″; M1 substitute x = 0, y = 1, y′ = 0; A1 y″ = 1; A1 positive, so a minimum.
Q6Implicit differentiation · Standard · Paper 2 style · 6 marks
  1. (a) The curve C has equation sin x + sin y = 1, for 0 < x < π/2, 0 < y < π/2. Find dy/dx at the point (π/6, π/6). [2]
  2. (b) Show that d²y/dx² at (π/6, π/6) is 2√3/3. [4]
Show answer
  1. (a) −1
    M1 cos x + cos y(dy/dx) = 0; A1 dy/dx = −cos x/cos y = −1.
  2. (b) 2√3/3
    M1 differentiate cos x + cos y·y′ = 0 again: −sin x − sin y·(y′)² + cos y·y″ = 0; M1 substitute values; A1 (√3/2)y″ = 1/2 + 1/2; A1 y″ = 2/√3 = 2√3/3.
Q7Implicit differentiation · Standard · Paper 2 style · 7 marks
  1. (a) The curve C has equation y³ + 3x²y = 4. Show that dy/dx = −2xy/(x² + y²). [3]
  2. (b) Find dy/dx and d²y/dx² at the point (1, 1), and describe what this suggests about the curve near this point. [4]
Show answer
  1. (a) Shown
    M1 3y²(dy/dx) + 6xy + 3x²(dy/dx) = 0; M1 collect terms; A1 result.
  2. (b) dy/dx = −1, d²y/dx² = 0; the curve is approximately straight there (possible point of inflection)
    B1 dy/dx = −1; M1 differentiate (x² + y²)y′ = −2xy: (2x + 2y y′)y′ + (x² + y²)y″ = −2y − 2x y′; M1 substitute: (2 − 2)(−1) + 2y″ = −2 + 2; A1 y″ = 0, with comment.
Q8Implicit differentiation · Challenge · Paper 2 style · 10 marks
  1. (a) The curve C has equation x² + xy + y² = 12. Find the coordinates of the points on C at which the tangent is parallel to the x-axis. [4]
  2. (b) Use d²y/dx² to determine the nature of each of these points. [4]
  3. (c) Find the coordinates of the points on C at which the tangent is parallel to the y-axis. [2]
Show answer
  1. (a) (2, −4) and (−2, 4)
    M1 dy/dx = −(2x + y)/(x + 2y); M1 dy/dx = 0 gives y = −2x; M1 substitute: 3x² = 12; A1 (2, −4) and (−2, 4).
  2. (b) (2, −4) is a minimum (d²y/dx² = 1/3); (−2, 4) is a maximum (d²y/dx² = −1/3)
    M1 2 + 2y′ + (x + 2y)y″ + 2(y′)² = 0; M1 with y′ = 0: y″ = −2/(x + 2y); A1 at (2, −4): y″ = 1/3, minimum; A1 at (−2, 4): y″ = −1/3, maximum.
  3. (c) (4, −2) and (−4, 2)
    M1 x + 2y = 0; A1 both points.
Q9Implicit differentiation · Challenge · Paper 2 style · 8 marks
  1. (a) The curve C has equation x³ + y³ = 6xy. Find the exact coordinates of the point on C, other than the origin, at which the tangent is parallel to the x-axis. [5]
  2. (b) Show that at this point d²y/dx² = −1. [3]
Show answer
  1. (a) (2^(4/3), 2^(5/3)) = (2∛2, 2∛4) ≈ (2.52, 3.17)
    M1 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx); A1 dy/dx = (2y − x²)/(y² − 2x); M1 dy/dx = 0 gives y = x²/2; M1 substitute: x³ + x⁶/8 = 3x³, x³ = 16; A1 x = 2∛2, y = 2∛4.
  2. (b) Shown
    M1 differentiate x² + y²y′ = 2y + 2xy′ again: 2x + 2y(y′)² + y²y″ = 4y′ + 2xy″; A1 with y′ = 0: y″ = −2x/(y² − 2x); A1 y² − 2x = 2^(10/3) − 2^(7/3) = 2^(7/3) = 2x, so y″ = −1 (a maximum).
Q10Parametric differentiation · Core · Paper 2 style · 4 marks
  1. (a) A curve is defined parametrically by x = t² + 1, y = t³ − 3t. Find dy/dx in terms of t. [2]
  2. (b) Find the coordinates of the stationary points of the curve. [2]
Show answer
  1. (a) dy/dx = (3t² − 3)/(2t)
    M1 (dy/dt)/(dx/dt); A1 result.
  2. (b) (2, −2) and (2, 2)
    M1 t = ±1; A1 t = 1 gives (2, −2), t = −1 gives (2, 2).
Q11Parametric differentiation · Core · Paper 2 style · 5 marks
  1. (a) An ellipse has parametric equations x = 2 cos θ, y = 3 sin θ. Find dy/dx in terms of θ. [2]
  2. (b) Find the equation of the tangent to the ellipse at the point where θ = π/4, giving your answer in the form ax + by = c√2. [3]
Show answer
  1. (a) dy/dx = −3 cos θ/(2 sin θ) = −(3/2) cot θ
    M1 dy/dθ = 3 cos θ, dx/dθ = −2 sin θ; A1 result.
  2. (b) 3x + 2y = 6√2
    M1 point (√2, 3√2/2); M1 gradient −3/2; A1 3x + 2y = 6√2.
Q12Parametric differentiation · Core · Paper 2 style · 5 marks
  1. (a) A curve has parametric equations x = 3t − 1, y = 2/t, t ≠ 0. Find dy/dx in terms of t. [2]
  2. (b) Find the equation of the normal to the curve at the point where t = 1. [3]
Show answer
  1. (a) −2/(3t²)
    M1 dy/dt = −2/t², dx/dt = 3; A1 −2/(3t²).
  2. (b) 2y = 3x − 2
    B1 point (2, 2); M1 normal gradient 3/2; A1 y − 2 = (3/2)(x − 2), i.e. 2y = 3x − 2.
Q13Parametric differentiation · Standard · Paper 2 style · 6 marks
  1. (a) A curve has parametric equations x = t + 1/t, y = t − 1/t, t > 1. Show that dy/dx = (t² + 1)/(t² − 1). [2]
  2. (b) Show that d²y/dx² = −4t³/(t² − 1)³ and find its value when t = 2. [4]
Show answer
  1. (a) Shown
    M1 dx/dt = 1 − 1/t², dy/dt = 1 + 1/t²; A1 multiply through by t².
  2. (b) −32/27
    M1 d/dt of dy/dx = −4t/(t² − 1)²; M1 divide by dx/dt = (t² − 1)/t²; A1 −4t³/(t² − 1)³; A1 −32/27.
Q14Parametric differentiation · Standard · Paper 2 style · 5 marks
  1. (a) A curve has parametric equations x = ln(1 + t²), y = 2 tan⁻¹ t, for t > 0. Show that dy/dx = 1/t. [2]
  2. (b) Find d²y/dx² in terms of t, and its value at the point where t = 1. [3]
Show answer
  1. (a) Shown
    M1 dx/dt = 2t/(1 + t²), dy/dt = 2/(1 + t²); A1 dy/dx = 1/t.
  2. (b) d²y/dx² = −(1 + t²)/(2t³); −1 when t = 1
    M1 d/dt(1/t) = −1/t²; M1 divide by 2t/(1 + t²); A1 −(1 + t²)/(2t³) = −1 at t = 1.
Q15Parametric differentiation · Standard · Paper 2 style · 6 marks
  1. (a) A curve has parametric equations x = eᵗ cos t, y = eᵗ sin t. Show that dy/dx = (sin t + cos t)/(cos t − sin t). [2]
  2. (b) Show that d²y/dx² = 2/(eᵗ(cos t − sin t)³) and find its value at t = 0. [4]
Show answer
  1. (a) Shown
    M1 product rule for both; A1 eᵗ cancels.
  2. (b) 2
    M1 quotient rule in t: numerator (cos t − sin t)² + (sin t + cos t)² = 2; A1 d/dt = 2/(cos t − sin t)²; M1 divide by eᵗ(cos t − sin t); A1 value 2 at t = 0.
Q16Parametric differentiation · Challenge · Paper 2 style · 8 marks
  1. (a) A curve has parametric equations x = t² + 2t, y = t³ − 12t. Find the values of t at the stationary points of the curve. [2]
  2. (b) Find d²y/dx² in terms of t. [3]
  3. (c) Find the coordinates of the stationary points and determine their nature. [3]
Show answer
  1. (a) t = 2 and t = −2
    M1 dy/dt = 3t² − 12 = 0; A1 t = ±2 (dx/dt ≠ 0 there).
  2. (b) d²y/dx² = (6t² + 12t + 24)/(2t + 2)³ = 3(t² + 2t + 4)/(4(t + 1)³)
    M1 dy/dx = (3t² − 12)/(2t + 2); M1 differentiate w.r.t. t and divide by dx/dt; A1 result.
  3. (c) (8, −16) minimum (d²y/dx² = 1/3); (0, 16) maximum (d²y/dx² = −3)
    A1 (8, −16) with 1/3 > 0, minimum; A1 (0, 16) with −3 < 0, maximum; B1 correct coordinates for both.
Q17Parametric differentiation · Challenge · Paper 2 style · 7 marks
  1. (a) A cycloid traced by a point on the rim of a bicycle wheel of radius a has parametric equations x = a(θ − sin θ), y = a(1 − cos θ), for 0 < θ < 2π. Show that dy/dx = cot(θ/2). [3]
  2. (b) Show that d²y/dx² = −1/(4a sin⁴(θ/2)), and deduce that d²y/dx² = −a/y². [4]
Show answer
  1. (a) Shown
    M1 dy/dx = sin θ/(1 − cos θ); M1 double-angle forms 2 sin(θ/2)cos(θ/2)/(2 sin²(θ/2)); A1 cot(θ/2).
  2. (b) Shown
    M1 d/dθ cot(θ/2) = −(1/2)cosec²(θ/2); M1 divide by a(1 − cos θ) = 2a sin²(θ/2); A1 −1/(4a sin⁴(θ/2)); A1 y = 2a sin²(θ/2) so y² = 4a² sin⁴(θ/2), giving −a/y².
Q18Parametric differentiation · Challenge · Paper 2 style · 7 marks
  1. (a) A curve has parametric equations x = t², y = t³. Find the equation of the tangent to the curve at the point P where t = 1. [3]
  2. (b) The tangent at P meets the curve again at Q. Find the coordinates of Q. [4]
Show answer
  1. (a) 2y = 3x − 1
    M1 dy/dx = 3t/2; A1 gradient 3/2 at (1, 1); A1 2y = 3x − 1.
  2. (b) (1/4, −1/8)
    M1 substitute: 2t³ = 3t² − 1; M1 2t³ − 3t² + 1 = (t − 1)²(2t + 1); A1 t = −1/2; A1 Q(1/4, −1/8).
Q19Maclaurin series · Core · Paper 2 style · 5 marks
  1. (a) Find the Maclaurin series for ln(1 + sin x) up to and including the term in x³. [5]
Show answer
  1. (a) x − x²/2 + x³/6
    M1 f′ = cos x/(1 + sin x), f′(0) = 1; M1 f″ = −1/(1 + sin x), f″(0) = −1; M1 f‴ = cos x/(1 + sin x)², f‴(0) = 1; M1 f(0) = 0 and use of Maclaurin formula; A1 x − x²/2 + x³/6.
Q20Maclaurin series · Core · Paper 2 style · 6 marks
  1. (a) Use differentiation to find the Maclaurin series for eˣ cos x up to and including the term in x³. [5]
  2. (b) Use your series to find an approximation to e^0.1 cos 0.1, giving your answer to 5 decimal places. [1]
Show answer
  1. (a) 1 + x − x³/3
    M1 f′ = eˣ(cos x − sin x); M1 f″ = −2eˣ sin x; M1 f‴ = −2eˣ(sin x + cos x); A1 f(0) = 1, f′(0) = 1, f″(0) = 0, f‴(0) = −2; A1 1 + x − (2/6)x³ = 1 + x − x³/3.
  2. (b) 1.09967
    B1 1 + 0.1 − 0.001/3 = 1.099667 → 1.09967.
Q21Maclaurin series · Standard · Paper 2 style · 7 marks
  1. (a) Given that f(x) = sec x, show that f″(x) = 2 sec³x − sec x. [3]
  2. (b) Hence find the Maclaurin series for sec x up to and including the term in x⁴. [4]
Show answer
  1. (a) Shown
    M1 f′ = sec x tan x; M1 f″ = sec x tan²x + sec³x; A1 use tan²x = sec²x − 1.
  2. (b) 1 + x²/2 + 5x⁴/24
    M1 f‴ = 6 sec³x tan x − sec x tan x, f‴(0) = 0; M1 f⁗ = 18 sec³x tan²x + 6 sec⁵x − sec x tan²x − sec³x, f⁗(0) = 5; A1 f(0) = 1, f″(0) = 1; A1 1 + x²/2 + 5x⁴/24.
Q22Maclaurin series · Standard · Paper 2 style · 5 marks
  1. (a) Find the Maclaurin series for tan⁻¹(1 + x) up to and including the term in x³. [5]
Show answer
  1. (a) π/4 + x/2 − x²/4 + x³/12
    B1 f(0) = π/4; M1 f′ = 1/(1 + (1 + x)²) = 1/(2 + 2x + x²), f′(0) = 1/2; M1 f″ = −(2 + 2x)/(2 + 2x + x²)², f″(0) = −1/2; M1 f‴(0) = 1/2; A1 π/4 + x/2 − x²/4 + x³/12.
Q23Maclaurin series · Standard · Paper 2 style · 8 marks
  1. (a) Find the Maclaurin series for eˣ√(1 + 2x) up to and including the term in x³. [5]
  2. (b) Use your series to estimate e^0.1 √1.2 correct to 6 decimal places, and find the percentage error of your estimate correct to 2 significant figures. [3]
Show answer
  1. (a) 1 + 2x + x² + 2x³/3
    M1 √(1 + 2x) = 1 + x − x²/2 + x³/2 + … (or differentiate directly); M1 multiply by 1 + x + x²/2 + x³/6; A1 x term 2x; A1 x² term 1/2 + 1 − 1/2 = 1; A1 x³ term 1/2 − 1/2 + 1/2 + 1/6 = 2/3.
  2. (b) Estimate 1.210667; percentage error ≈ 0.0010 % (an overestimate)
    M1 1 + 0.2 + 0.01 + 0.002/3; A1 1.210667; B1 true value 1.210654, error ≈ 1.26 × 10⁻⁵, i.e. 0.0010 %.
Q24Maclaurin series · Challenge · Paper 2 style · 8 marks
  1. (a) Given f(x) = ln(cos x), for −π/2 < x < π/2, find the Maclaurin series for f(x) up to and including the term in x⁴. [6]
  2. (b) Hence, without further differentiation, find the Maclaurin series for tan x up to and including the term in x³. [2]
Show answer
  1. (a) −x²/2 − x⁴/12
    M1 f′ = −tan x, f″ = −sec²x, f‴ = −2 sec²x tan x; M1 f⁗ = −4 sec²x tan²x − 2 sec⁴x; A1 f(0) = 0, f′(0) = 0, f″(0) = −1, f‴(0) = 0, f⁗(0) = −2; M1 Maclaurin formula; A1 −x²/2 − 2x⁴/24; A1 = −x²/2 − x⁴/12.
  2. (b) x + x³/3
    M1 tan x = −d/dx ln(cos x); A1 −(−x − x³/3) = x + x³/3.
Q25Maclaurin series · Challenge · Paper 2 style · 8 marks
  1. (a) Given that y = sin⁻¹x, show that (1 − x²)(d²y/dx²) − x(dy/dx) = 0. [3]
  2. (b) By differentiating this result three more times, find the Maclaurin series for sin⁻¹x up to and including the term in x⁵. [5]
Show answer
  1. (a) Shown
    M1 dy/dx = (1 − x²)^(−1/2), so √(1 − x²)·dy/dx = 1; M1 differentiate: √(1 − x²)y″ − x(1 − x²)^(−1/2)y′ = 0; A1 multiply by √(1 − x²).
  2. (b) x + x³/6 + 3x⁵/40
    M1 (1 − x²)y‴ − 3xy″ − y′ = 0; M1 (1 − x²)y⁗ − 5xy‴ − 4y″ = 0, and once more (1 − x²)y⁽⁵⁾ − 7xy⁗ − 9y‴ = 0; A1 y(0) = 0, y′(0) = 1, y″(0) = 0, y‴(0) = 1, y⁗(0) = 0, y⁽⁵⁾(0) = 9; M1 Maclaurin formula; A1 x + x³/6 + 9x⁵/120 = x + x³/6 + 3x⁵/40.
Q26Derivatives of hyperbolic & inverse functions · Core · Paper 2 style · 5 marks
  1. (a) Find dy/dx when y = sinh⁻¹(3x). [2]
  2. (b) Find dy/dx when y = cosh(x²). [1]
  3. (c) Find dy/dx when y = tanh⁻¹(x/2), for |x| < 2, simplifying your answer. [2]
Show answer
  1. (a) 3/√(9x² + 1)
    M1 chain rule with d/du sinh⁻¹u = 1/√(u² + 1); A1.
  2. (b) 2x sinh(x²)
    B1
  3. (c) 2/(4 − x²)
    M1 (1/2)/(1 − x²/4); A1 2/(4 − x²).
Q27Derivatives of hyperbolic & inverse functions · Core · Paper 2 style · 4 marks
  1. (a) Differentiate y = x² tanh 3x with respect to x. [2]
  2. (b) Find the exact gradient of the curve y = cos⁻¹(x/2) at the point where x = 1. [2]
Show answer
  1. (a) 2x tanh 3x + 3x² sech²3x
    M1 product rule; A1.
  2. (b) −1/√3 = −√3/3
    M1 dy/dx = −1/√(4 − x²); A1 −1/√3.
Q28Derivatives of hyperbolic & inverse functions · Standard · Paper 2 style · 6 marks
  1. (a) By writing x = cosh y, prove that d/dx (cosh⁻¹x) = 1/√(x² − 1), for x > 1. [3]
  2. (b) Find the gradient of the curve y = x cosh⁻¹x at the point where x = 2, giving your answer in exact form and correct to 3 decimal places. [3]
Show answer
  1. (a) Shown
    M1 dx/dy = sinh y; M1 sinh y = √(cosh²y − 1) = √(x² − 1) (positive since y > 0); A1 dy/dx = 1/√(x² − 1).
  2. (b) ln(2 + √3) + 2/√3 ≈ 2.472
    M1 product rule: cosh⁻¹x + x/√(x² − 1); M1 cosh⁻¹2 = ln(2 + √3); A1 ln(2 + √3) + 2√3/3 = 2.472.
Q29Derivatives of hyperbolic & inverse functions · Standard · Paper 2 style · 6 marks
  1. (a) Show that d/dx [tan⁻¹(eˣ)] = (1/2) sech x. [3]
  2. (b) Hence find the exact value of ∫₀^(ln 3) sech x dx. [3]
Show answer
  1. (a) Shown
    M1 chain rule: eˣ/(1 + e²ˣ); M1 divide numerator and denominator by eˣ: 1/(e⁻ˣ + eˣ); A1 = 1/(2 cosh x) = (1/2) sech x.
  2. (b) 2 tan⁻¹3 − π/2 ≈ 0.927
    M1 ∫ sech x dx = 2 tan⁻¹(eˣ); M1 limits: 2 tan⁻¹3 − 2 tan⁻¹1; A1 2 tan⁻¹3 − π/2 ≈ 0.927.
Q30Derivatives of hyperbolic & inverse functions · Challenge · Paper 2 style · 7 marks
  1. (a) The curve C has equation y = 3 cosh x − sinh x. Show that C has exactly one stationary point and find its exact coordinates. [5]
  2. (b) Determine the nature of the stationary point. [2]
Show answer
  1. (a) (½ ln 2, 2√2)
    M1 dy/dx = 3 sinh x − cosh x = 0; M1 tanh x = 1/3 (only one solution since tanh is one-one); M1 x = tanh⁻¹(1/3) = ½ ln((1 + 1/3)/(1 − 1/3)) = ½ ln 2; M1 cosh x = 3/(2√2), sinh x = 1/(2√2); A1 y = 9/(2√2) − 1/(2√2) = 2√2.
  2. (b) Minimum
    M1 d²y/dx² = 3 cosh x − sinh x = y; A1 at the point d²y/dx² = 2√2 > 0, so a minimum.

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Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Implicit differentiation 345/9
Parametric differentiation 334/9
Maclaurin series 266/9
Derivatives of hyperbolic & inverse functions 144/9

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