Free worked solutions · 2026 Edition

Chapter 14: Differentiation worked solutions

Worked solutions to the practice questions in Chapter 14 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[6 marks]

Let f(x) = 4x^2 - x^3. Starting from the definition f'(x) = \displaystyle\lim_{h \to 0}\left( \dfrac{f(x + h) - f(x)}{h} \right), expand f(x + h), simplify the quotient and hence obtain an expression for f'(x).

[6]

Show solutionHide solution

f(x + h) = 4(x + h)^2 - (x + h)^3

4(x + h)^2 = 4x^2 + 8xh + 4h^2

(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3

f(x + h) - f(x) = 8xh + 4h^2 - 3x^2h - 3xh^2 - h^3

\dfrac{f(x + h) - f(x)}{h} = 8x + 4h - 3x^2 - 3xh - h^2

f'(x) = \displaystyle\lim_{h \to 0} \left( 8x - 3x^2 + 4h - 3xh - h^2 \right) = 8x - 3x^2

Note: do not award the last two A1 if h = 0 is substituted before cancelling. The \displaystyle\lim_{h \to 0} may be omitted from intermediate lines but must appear in the conclusion.

Final answer

f'(x)=8x-3x^2

Question 2

[10 marks]

The table shows some values of the functions f and g and of their derivatives f' and g'.

x	1	2	3
f(x)	6	2	-4
f'(x)	-1	3	5
g(x)	3	2	-1
g'(x)	4	-1	2

(a) Let h(x) = f(x)\,g(x). Find h'(2).

[2]

(b) Let q(x) = \dfrac{f(x)}{g(x)}. Find q'(2).

[2]

(c) Let r(x) = f(g(x)). Find r'(1).

[2]

(d) Let p(x) = \ln\left(f(x)\right). Find p'(2).

[2]

(e) Let w(x) = \left(g(x)\right)^3. Find w'(3).

[2]

Show solutionHide solution

(a) h'(2) = f'(2)g(2) + f(2)g'(2) = 3 \times 2 + 2 \times (-1) (M1)

= 4

(b) q'(2) = \dfrac{g(2)f'(2) - f(2)g'(2)}{\left(g(2)\right)^2} = \dfrac{2 \times 3 - 2 \times (-1)}{4} (M1)

= 2

(c) r'(1) = f'\left(g(1)\right) \times g'(1) = f'(3) \times 4 (M1)

= 5 \times 4 = 20

(d) p'(2) = \dfrac{f'(2)}{f(2)} (valid as f(2) = 2 > 0) (M1)

= \dfrac{3}{2}

(e) w'(3) = 3\left(g(3)\right)^2 g'(3) = 3 \times 1 \times 2 (M1)

= 6

Final answer

(a) 4
(b) 2
(c) 20
(d) \frac32
(e) 6

Question 3

[6 marks]

The line y = 4x + c is a tangent to the curve y = x^3 - 8x.

Find the possible values of c.

[6]

Show solutionHide solution

\dfrac{dy}{dx} = 3x^2 - 8

at the point of contact the gradients are equal: 3x^2 - 8 = 4

x^2 = 4 \Rightarrow x = 2 or x = -2

points of contact (2,\ -8) and (-2,\ 8)

c = y - 4x: at (2,\ -8), c = -8 - 8 = -16

at (-2,\ 8), c = 8 + 8 = 16

Note: the discriminant method does not apply here because the equation x^3 - 8x = 4x + c is a cubic.

Final answer

c=-16 or c=16

Question 4

[7 marks]

The curve C has equation y = \operatorname{cosec}\,2x, for 0 < x < \dfrac{\pi}{2}. The point P on C has x-coordinate \dfrac{\pi}{8}.

(a) Show that the y-coordinate of P is \sqrt{2}, and find the gradient of C at P.

[3]

(b) Find the equation of the tangent to C at P, in the form y = mx + c.

[2]

(c) Find the equation of the normal to C at P.

[2]

Show solutionHide solution

(a) \mathrm{cosec}\,\dfrac{\pi}{4} = \dfrac{1}{\sin\frac{\pi}{4}} = \sqrt{2}

\dfrac{dy}{dx} = -2\,\mathrm{cosec}\,2x\cot 2x M1A1

at x = \dfrac{\pi}{8}: -2 \times \sqrt{2} \times 1 = -2\sqrt{2}

(b) y - \sqrt{2} = -2\sqrt{2}\left( x - \dfrac{\pi}{8} \right) (M1)

y = -2\sqrt{2}\,x + \sqrt{2} + \dfrac{\sqrt{2}\,\pi}{4}

(c) gradient of the normal = \dfrac{1}{2\sqrt{2}} = \dfrac{\sqrt{2}}{4} (M1)

y - \sqrt{2} = \dfrac{\sqrt{2}}{4}\left( x - \dfrac{\pi}{8} \right)

Note: Accept any correct equivalent form in (c). FT their gradient from (a).

Final answer

(a) gradient -2\sqrt2
(b) y=-2\sqrt2\,x+\sqrt2+\frac{\sqrt2\pi}{4}
(c) y-\sqrt2=\frac{\sqrt2}{4}\left(x-\frac{\pi}{8}\right)

Question 5

[11 marks]

Let f(x) = x^3 + px^2 + qx + 1, where p and q are constants. The graph of f has a point of inflexion at (2,\ -3).

(a) Find f''(x), and hence find the value of p.

[3]

(b) Find the value of q.

[2]

(c) Show that the point of inflexion is not a stationary point.

[2]

(d) Find the exact x-coordinates of the two stationary points, and determine the nature of each.

[4]

Show solutionHide solution

(a) f'(x) = 3x^2 + 2px + q, f''(x) = 6x + 2p

f''(2) = 0 \Rightarrow 12 + 2p = 0

p = -6

(b) f(2) = -3 \Rightarrow 8 - 24 + 2q + 1 = -3

q = 6

(c) f'(x) = 3x^2 - 12x + 6, so f'(2) = 12 - 24 + 6

= -6 \ne 0, so it is not a stationary point

(d) 3x^2 - 12x + 6 = 0 \Rightarrow x^2 - 4x + 2 = 0

x = 2 \pm \sqrt{2}

f''(2 - \sqrt{2}) = 6(2 - \sqrt{2}) - 12 = -6\sqrt{2} < 0: local maximum at x = 2 - \sqrt{2}

f''(2 + \sqrt{2}) = 6\sqrt{2} > 0: local minimum at x = 2 + \sqrt{2}

Final answer

(a) f''(x)=6x+2p, p=-6
(b) q=6
(c) f'(2)=-6\neq0
(d) maximum at x=2-\sqrt2, minimum at x=2+\sqrt2

Question 6

[7 marks]

A curve has equation x = y^3 + 2y.

(a) Find \dfrac{dx}{dy}, and hence show that \dfrac{dy}{dx} = \dfrac{1}{3y^2 + 2}.

[2]

(b) The point P on the curve has y-coordinate 1. Find the x-coordinate of P and the equation of the tangent to the curve at P.

[3]

(c) Explain why the gradient of the curve is never greater than \tfrac{1}{2}, and write down the coordinates of the point where it equals \tfrac{1}{2}.

[2]

Show solutionHide solution

(a) \dfrac{dx}{dy} = 3y^2 + 2

\dfrac{dy}{dx} = \dfrac{1}{dx/dy} = \dfrac{1}{3y^2 + 2} M1AG

Note: Accept differentiating x = y^3 + 2y implicitly with respect to x: 1 = \left( 3y^2 + 2 \right)\dfrac{dy}{dx}.

(b) x = 1 + 2 = 3

gradient \dfrac{1}{5}

y - 1 = \dfrac{1}{5}(x - 3) \left( y = \dfrac{1}{5}x + \dfrac{2}{5} \right)

(c) 3y^2 + 2 \ge 2 for all y, so \dfrac{dy}{dx} = \dfrac{1}{3y^2 + 2} \le \dfrac{1}{2}

equality when y = 0: the point (0,\ 0)

Final answer

(b) x=3; y=\frac15x+\frac25
(c) greatest gradient \frac12, at (0,0)

Question 7

[6 marks]

A hailstone is modelled as a sphere of radius r mm. As it falls through a cloud, ice forms on it so that its volume increases at a constant rate of 50\ \text{mm}^3 per minute.

(a) Find the exact rate of increase of r when r = 3.

[3]

(b) Find the rate of increase of the surface area of the hailstone at the same instant.

[3]

Show solutionHide solution

(a) V = \dfrac{4}{3}\pi r^3, \dfrac{dV}{dr} = 4\pi r^2

50 = 4\pi(3)^2\dfrac{dr}{dt}

\dfrac{dr}{dt} = \dfrac{50}{36\pi} = \dfrac{25}{18\pi} mm per minute (0.442)

(b) S = 4\pi r^2, \dfrac{dS}{dr} = 8\pi r

\dfrac{dS}{dt} = 8\pi(3) \times \dfrac{25}{18\pi}

= \dfrac{100}{3}\ \text{mm}^2 per minute (33.3)

Note: FT from (a) in (b).

Final answer

(a) \frac{25}{18\pi}\approx0.442 mm per minute
(b) \frac{100}{3}\approx33.3 mm^2 per minute

Question 8

[8 marks]

Ravi writes

\begin{aligned}\lim_{x \to 0} \frac{1 - \cos 2x}{x^2 + 3x} &= \lim_{x \to 0} \frac{2\sin 2x}{2x + 3}\\ &= \lim_{x \to 0} \frac{4\cos 2x}{2} = 2.\end{aligned}

(a) Identify the error in Ravi's working, and find the correct value of the limit.

[3]

(b) Use l'Hôpital's rule to find

\lim_{x \to 0} \frac{1 - \cos 2x}{x^2 + 3x^3}.

[5]

Show solutionHide solution

(a) the first step is valid: 1 - \cos 0 = 0 and 0, the form \dfrac{0}{0}

after it, \dfrac{2\sin 2x}{2x + 3} \to \dfrac{0}{3}, which is not indeterminate, so the second application of the rule is not allowed

correct value: \dfrac{0}{3} = 0

(b) 1 - \cos 0 = 0 and 0: the form \dfrac{0}{0}

\displaystyle\lim_{x \to 0} \dfrac{2\sin 2x}{2x + 9x^2} M1A1

\sin 0 = 0 and 0: \dfrac{0}{0} again (M1)

\displaystyle\lim_{x \to 0} \dfrac{4\cos 2x}{2 + 18x} = \dfrac{4}{2} = 2

Note: award M0 if the quotient rule is applied to the whole fraction instead of differentiating the numerator and the denominator separately. Do not award the final A1 if the form was not checked at a stage where it was needed. The R1 is for the first form; the (M1) for a repeated application is implied only by a correctly stated form (\dfrac{0}{0} or \dfrac{\infty}{\infty}) at that stage.

Final answer

(a) after one step the denominator tends to 3, so the form is no longer \frac00 and the rule must stop; the limit is 0
(b) 2

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 9

[6 marks]

Let f(x) = x^4 - 4x^3.

(a) Find f'(x) and the x-coordinates of the stationary points of the graph of f.

[2]

(b) Determine the nature of each stationary point, justifying your answers.

[4]

Show solutionHide solution

(a) f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3).

[1]

Stationary points at x = 0 and x = 3.

[1]

(b) f''(x) = 12x^2 - 24x.

[1]

f''(3) = 36 > 0, so (3, -27) is a local minimum.

[1]

f''(0) = 0, so the second derivative test is inconclusive at x = 0.

[1]

f'(x) = 4x^2(x - 3) < 0 on both sides of x = 0 (close to 0): the gradient does not change sign, so (0, 0) is a horizontal point of inflexion, neither a maximum nor a minimum.

[1]

Final answer

(a) f'(x)=4x^2(x-3); x=0, x=3
(b) (3,-27) minimum; (0,0) stationary point of inflexion

Question 10

[6 marks]

A rumour starts in a school of 1200 pupils. The number of pupils who have heard the rumour t days after it starts is modelled by

N(t) = \frac{1200}{1 + 23e^{-0.5t}}, \qquad t \geq 0.

The graph of N has a point of inflexion.

(a) Use your GDC to find the value of t at which the point of inflexion occurs.

[2]

(b) Find the value of N at this point.

[1]

(c) Find the rate at which the number of pupils who have heard the rumour is increasing at this time.

[1]

(d) Describe, in context, how the rate at which the rumour spreads changes before and after this time.

[2]

Show solutionHide solution

(a) The inflexion is where N'(t) has its maximum. GDC: define f_2(x) = \dfrac{\mathrm{d}}{\mathrm{d}x}f_1(x) and use Analyze Graph \to Maximum on the graph of N'.

[1]

t = 6.27 days (3 s.f.; unrounded 6.2710\ldots).

[1]

(b) N(6.2710) = 600, half of the school.

[1]

(c) N'(6.2710) = 150 pupils per day (GDC numerical derivative).

[1]

(d) Before t = 6.27 the graph is concave up: the rumour spreads faster and faster, reaching its fastest spread of 150 pupils per day.

[1]

After t = 6.27 the graph is concave down: the rate falls (fewer pupils are left who have not heard it), although N still increases towards 1200.

[1]

Final answer

(a) t=6.27
(b) N=600
(c) 150 pupils per day
(d) the rate rises to a maximum at this time, then falls

Question 11

[7 marks]

A cargo ship makes a voyage of 1500 km at a constant speed of v km h^{-1}. The fuel cost is 0.004v^3 dollars per hour, and all other costs total 600 dollars per hour. The total cost of the voyage is C dollars.

(a) Show that C = 6v^2 + \dfrac{900\,000}{v}.

[2]

(b) Find \dfrac{dC}{dv}, and hence find the speed that minimises C, justifying that it is a minimum.

[4]

(c) Find the minimum cost of the voyage.

[1]

Show solutionHide solution

(a) The voyage takes \dfrac{1500}{v} hours, and the cost per hour is 0.004v^3 + 600.

[1]

C = \left(0.004v^3 + 600\right)\dfrac{1500}{v} = 6v^2 + \dfrac{900\,000}{v}. [1]

(b) \dfrac{\mathrm{d}C}{\mathrm{d}v} = 12v - \dfrac{900\,000}{v^2}.

[1]

12v - \dfrac{900\,000}{v^2} = 0 \Rightarrow v^3 = 75\,000.

[1]

v = \sqrt[3]{75\,000} = 42.2 km h^{-1} (3 s.f.; unrounded 42.1716\ldots). Check on the GDC: graph y = 6x^2 + \frac{900\,000}{x} and use Analyze Graph \to Minimum.

[1]

\dfrac{\mathrm{d}^2C}{\mathrm{d}v^2} = 12 + \dfrac{1\,800\,000}{v^3} > 0 for v > 0, so this is a minimum.

[1]

(c) C = 6(42.1716)^2 + \dfrac{900\,000}{42.1716} = 32\,012\ldots \approx 32\,000 dollars (3 s.f.).

[1]

Final answer

(b) v=\sqrt[3]{75\,000}\approx42.2 km h^{-1}
(c) C\approx32\,000 dollars

Question 12

[8 marks]

The point A has coordinates (0, 3.2), and the point P(x, x^2) lies on the graph of y = x^2.

(a) Show that AP^2 = x^4 - 5.4x^2 + 10.24.

[2]

(b) Find the possible x-coordinates of P when the distance AP is a minimum. Justify that your values give a minimum.

[4]

(c) Find the minimum value of AP.

[2]

Show solutionHide solution

(a) \mathrm{AP}^2 = (x - 0)^2 + (x^2 - 3.2)^2.

[1]

= x^2 + x^4 - 6.4x^2 + 10.24 = x^4 - 5.4x^2 + 10.24. [1]

(b) \mathrm{AP} > 0, so \mathrm{AP} is least when \mathrm{AP}^2 is least. Let D = x^4 - 5.4x^2 + 10.24: \dfrac{\mathrm{d}D}{\mathrm{d}x} = 4x^3 - 10.8x.

[1]

4x^3 - 10.8x = 4x(x^2 - 2.7) = 0 \Rightarrow x = 0 or x^2 = 2.7.

[1]

D(0) = 10.24 whereas D(\sqrt{2.7}) = 7.29 - 14.58 + 10.24 = 2.95. So x = 0 is not the minimum, and the minimum occurs at x = \pm\sqrt{2.7} = \pm 1.64 (3 s.f.; unrounded 1.6432\ldots). GDC check: graph y = \sqrt{x^2 + (x^2 - 3.2)^2} and use Analyze Graph \to Minimum. [1]

[1]

(c) Minimum \mathrm{AP}^2 = 2.95.

[1]

Minimum \mathrm{AP} = \sqrt{2.95} = 1.72 (3 s.f.; unrounded 1.7176\ldots).

[1]

Final answer

(b) x=\pm1.64
(c) AP_{\min}=1.72

Question 13

[6 marks]

A particle moves in a straight line with displacement s(t) = 2t^3 - 14t^2 + 21t metres from the origin at time t seconds, for t \geq 0.

(a) Find the velocity and the acceleration of the particle when t = 1, and state whether the particle is speeding up or slowing down at that time.

[3]

(b) Find the values of t for which the particle is speeding up.

[3]

Show solutionHide solution

(a) v = 6t^2 - 28t + 21, so v(1) = -1 m s^{-1}.

[1]

a = 12t - 28, so a(1) = -16 m s^{-2}.

[1]

The velocity and the acceleration are both negative (same sign), so the particle is speeding up (moving in the negative direction ever faster).

[1]

(b) Speeding up when v and a have the same sign. GDC: solve 6t^2 - 28t + 21 = 0 to get t = 0.939 and t = 3.73 (unrounded 0.93890\ldots and 3.72777\ldots); a = 0 at t = \dfrac{7}{3} = 2.33.

[1]

v < 0 for 0.939 < t < 3.73 and v > 0 otherwise; a < 0 for t < 2.33 and a > 0 for t > 2.33.

Both negative for 0.939 < t < 2.33.

[1]

Both positive for t > 3.73.

[1]

Final answer

(a) v=-1, a=-16: speeding up
(b) 0.939<t<2.33 and t>3.73

Question 14

[8 marks]

A spherical tank of radius 5 m is being drained. When the depth of liquid in the tank is h metres, 0 \leq h \leq 10, the volume of liquid is

V = \frac{\pi}{3}h^2(15 - h)\ \text{m}^3.

Liquid leaves the tank at a constant rate of 0.8 m^3 per minute.

(a) Show that \dfrac{dV}{dh} = \pi h(10 - h).

[2]

(b) Find the rate of change of the depth when h = 4.

[3]

(c) Find the depth at which the level is falling most slowly. Justify your answer.

[3]

Show solutionHide solution

; (a) V = \dfrac{\pi}{3}\left(15h^2 - h^3\right).

[1]

\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{\pi}{3}\left(30h - 3h^2\right) = \pi h(10 - h). [1]

(b) \dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}h}\times\dfrac{\mathrm{d}h}{\mathrm{d}t}.

[1]

-0.8 = \pi(4)(6)\dfrac{\mathrm{d}h}{\mathrm{d}t}.

[1]

\dfrac{\mathrm{d}h}{\mathrm{d}t} = -\dfrac{0.8}{24\pi} = -0.0106 m per minute (3 s.f.).

[1]

(c) \left|\dfrac{\mathrm{d}h}{\mathrm{d}t}\right| = \dfrac{0.8}{\pi h(10 - h)}, which is smallest when h(10 - h) is greatest.

[1]

h(10 - h) = 10h - h^2 has derivative 10 - 2h = 0 at h = 5.

[1]

10h - h^2 is a downward-opening parabola, so h = 5 gives its maximum: the level falls most slowly at depth 5 m (the widest part of the tank), at \dfrac{0.8}{25\pi} = 0.0102 m per minute.

[1]

Final answer

(b) -0.0106 m per minute
(c) h=5 m

Question 15

[8 marks]

The curve C has equation xe^y + y^3 = 8. The point P on C has x-coordinate 1 and y > 0.

(a) Use your GDC to find the y-coordinate of P.

[2]

(b) Show that \dfrac{dy}{dx} = -\dfrac{e^y}{xe^y + 3y^2}.

[3]

(c) Find the gradient of C at P, and the equation of the tangent to C at P.

[3]

Show solutionHide solution

; (a) At x = 1: e^y + y^3 = 8.

[1]

GDC: solve for y > 0 (or graph y = e^x + x^3 and y = 8): y = 1.51 (3 s.f.; unrounded 1.5127\ldots).

[1]

(b) Differentiate with respect to x (product rule on xe^y): e^y + xe^y\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0. [1]

[1]

\left(xe^y + 3y^2\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} = -e^y, so \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{e^y}{xe^y + 3y^2}. [1]

(c) At (1, 1.5127): \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{e^{1.5127}}{e^{1.5127} + 3(1.5127)^2}.

[1]

= -0.398 (3 s.f.; unrounded -0.39803\ldots).

[1]

Tangent: y - 1.5127 = -0.398(x - 1), i.e. y = -0.398x + 1.91.

[1]

Final answer

(a) y=1.51
(c) gradient -0.398; y=-0.398x+1.91

Question 16

[6 marks]

Use l'Hôpital's rule to find each limit. State the form of the limit at each stage.

(a) \displaystyle\lim_{x \to 0}\frac{1 - \cos 3x}{x^2}

[3]

(b) \displaystyle\lim_{x \to \infty}\frac{x^2}{e^{3x}}

[3]

Show solutionHide solution

; (a) The form is \dfrac{0}{0}: \lim_{x \to 0}\dfrac{3\sin 3x}{2x}.

[1]

This is still \dfrac{0}{0}: \lim_{x \to 0}\dfrac{9\cos 3x}{2}.

[1]

= \dfrac{9}{2}.

[1]

(b) The form is \dfrac{\infty}{\infty}: \lim_{x \to \infty}\dfrac{2x}{3e^{3x}}.

[1]

Still \dfrac{\infty}{\infty}: \lim_{x \to \infty}\dfrac{2}{9e^{3x}}.

[1]

= 0.

[1]

Final answer

(a) \frac92
(b) 0

Question 17

[7 marks]

Let y = x\ln x - \dfrac{x^2}{4}, for x > 0. The line L has equation y = \dfrac{x}{2}.

(a) Find \dfrac{dy}{dx}.

[2]

(b) The tangent to the graph is parallel to L at two points. Use your GDC to find the x-coordinates of these points.

[3]

(c) Find the y-coordinates of these two points.

[2]

Show solutionHide solution

(a) \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ln x + x\cdot\dfrac{1}{x} - \dfrac{x}{2} = \ln x + 1 - \dfrac{x}{2}. [1]

[1]

(b) Parallel to L means \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}: \ln x + 1 - \dfrac{x}{2} = \dfrac{1}{2}.

[1]

GDC: graph y = \ln x + 1 - \frac{x}{2} and y = 0.5 (or solve) and find the intersections.

x = 1 and x = 3.51 (3 s.f.; unrounded 3.5129\ldots). [1]

[1]

(c) x = 1: y = 0 - \dfrac{1}{4} = -0.25.

[1]

x = 3.5129: y = 3.5129\ln(3.5129) - \dfrac{3.5129^2}{4} = 1.33 (3 s.f.).

[1]

Final answer

(a) \frac{dy}{dx}=\ln x+1-\frac x2
(b) x=1, x=3.51
(c) y=-0.25, y=1.33

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 18

[4 marks]

Let h(x) = 3^{\,x^{2} - 4x}, where x \in \mathbb{R}. Find the exact value of h'(1).

[4]

Show solutionHide solution

Using \dfrac{\mathrm{d}}{\mathrm{d}x}\left(a^{u}\right) = u'\,a^{u}\ln a (or rewriting as \mathrm{e}^{(x^2-4x)\ln 3}):

h'(x) = (2x - 4)\ln 3\cdot 3^{\,x^2 - 4x}

A1A1

At x = 1: the exponent is 1 - 4 = -3, so

h'(1) = (-2)\ln 3\cdot 3^{-3} = \mathbf{-\frac{2\ln 3}{27}}

A1[4]

Final answer

h'(1)=-\dfrac{2\ln3}{27}

Question 19

[5 marks]

A spherical snowball is melting so that its surface area is decreasing at a constant rate of 6\ \text{cm}^{2}\text{s}^{-1}.

Determine the rate at which its radius r cm is changing at the instant when the volume of the snowball is 36\pi\ \text{cm}^{3}.

[5]

Show solutionHide solution

The volume and surface area of a sphere are

V = \frac{4}{3}\pi r^{3}, \qquad S = 4\pi r^{2} \qquad \textbf{(M1)}

When V = 36\pi: r^{3} = 27, so r = 3.

Connecting the rates by the chain rule:

\frac{\mathrm{d}S}{\mathrm{d}t} = \frac{\mathrm{d}S}{\mathrm{d}r}\times\frac{\mathrm{d}r}{\mathrm{d}t} = 8\pi r\,\frac{\mathrm{d}r}{\mathrm{d}t}

Substituting \dfrac{\mathrm{d}S}{\mathrm{d}t} = -6 and r = 3:

-6 = 24\pi\,\frac{\mathrm{d}r}{\mathrm{d}t} \qquad \textbf{A1}

\frac{\mathrm{d}r}{\mathrm{d}t} = -\frac{1}{4\pi}\ \text{cm s}^{-1}\ (\approx -0.0796) \qquad \textbf{A1}

So the radius is decreasing at \dfrac{1}{4\pi}\ \text{cm s}^{-1}.

[5]

Final answer

\frac{dr}{dt}=-\frac{1}{4\pi}\approx-0.0796 cm s^{-1}

Question 20

[6 marks]

A differentiable function f satisfies f'(x) = 4 + 5x - 3e^{x} for every real x.

(a) For which real numbers x is f increasing?

[3]

(b) Determine where the graph of f is concave down, stating your answer as an inequality in x.

[3]

Show solutionHide solution

(a) f is increasing when f'(x) > 0. The equation f'(x) = 0 has roots (from a GDC)

x = -0.396328\ldots \quad \text{and} \quad x = 1.20793\ldots

Testing a value in each region — f'(-3) \approx -11.1, f'(0) = 1, f'(3) \approx -41.3 — shows that f' is positive only between the roots:

-0.396 < x < 1.21

[3]

(b) Concave down requires f''(x) < 0, where f''(x) = 5 - 3\mathrm{e}^{x}:

3\mathrm{e}^{x} > 5 \implies x > \ln\frac{5}{3}

x > 0.511

[3]

Final answer

(a) -0.396<x<1.21
(b) x>\ln\frac53\approx0.511

Question 21

[7 marks]

Let f(x) = \dfrac{(k-3x)^2}{(x+2)^3} for x \neq -2, where k is a positive constant.

(a) Differentiate f, giving your answer in terms of k and x.

[3]

At x = 2 the tangent to the curve y=f(x) is inclined at 10^{\circ} to the positive x-axis.

(b) Calculate both possible values of k.

[4]

Show solutionHide solution

(a)

Using the quotient rule with the chain rule on the numerator:

f'(x) = \frac{-6(k-3x)(x+2)^3 - 3(k-3x)^2(x+2)^2}{(x+2)^6}

Taking out the common factor -3(k-3x)(x+2)^2 from the numerator:

f'(x) = \frac{-3(k-3x)\left[2(x+2) + (k-3x)\right]}{(x+2)^4} = \frac{-3(k-3x)(k-x+4)}{(x+2)^4} \qquad \textbf{A1}

[3]

(b)

Recognising that f'(2) = \tan 10^{\circ}\ (= 0.176326\ldots): (M1)

\frac{-3(k-6)(k+2)}{256} = \tan 10^{\circ} \qquad \textbf{A1}

Solving this equation: (M1)

k = 1.02356\ldots,\ 2.97643\ldots

k = 1.02,\ 2.98

[4]

Final answer

(a) f'(x)=\frac{-3(k-3x)(k-x+4)}{(x+2)^4}
(b) k=1.02 or k=2.98

Question 22

[9 marks]

Let g(x) = \ln(x^2 - 4x) for all x > 4.

The curve y = g(x) meets the x-axis at the point A = (a, 0). The tangent T to the curve at a point B has slope \dfrac{1}{3}.

(a) Determine, in exact form, the value of a.

[3]

(b) Determine, in exact form, the x-coordinate of B.

[6]

Show solutionHide solution

(a)

At A, g(x) = 0:

x^2 - 4x = 1 \implies x^2 - 4x - 1 = 0 \implies x = 2 \pm \sqrt{5}

The domain requires x > 4, and 2 - \sqrt{5} < 0, so a = 2 + \sqrt{5}

[3]

(b)

The chain rule gives

g'(x) = \frac{2x - 4}{x^2 - 4x}

At B the slope is \dfrac{1}{3}:

3(2x - 4) = x^2 - 4x \implies x^2 - 10x + 12 = 0

By the quadratic formula,

x = 5 \pm \sqrt{13}

The root 5 - \sqrt{13} \approx 1.39 is not greater than 4, so it is rejected. Hence x = 5 + \sqrt{13} (\approx 8.61)

[6]

Final answer

(a) a=2+\sqrt5
(b) x=5+\sqrt{13}

Question 23

[14 marks]

The ellipse E has equation \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1.

A rectangle is drawn with its sides parallel to the coordinate axes and its four vertices on E. One vertex, Q(x, y), lies in the first quadrant, so the vertices are (\pm x,\ \pm y).

(a) Show that y = \dfrac{3}{5}\sqrt{25 - x^2}.

[1]

(b) Express the rectangle's area A as a function of x alone.

[3]

(c) Show that \dfrac{dA}{dx} = \dfrac{12\left(25 - 2x^2\right)}{5\sqrt{25 - x^2}}.

[4]

(d) Determine the perimeter of the rectangle that has the largest area, writing it as p\sqrt{2} with p a positive integer.

[6]

Show solutionHide solution

(a)

\frac{y^2}{9} = 1 - \frac{x^2}{25} = \frac{25 - x^2}{25} \implies y^2 = \frac{9}{25}\left(25 - x^2\right)

Since Q is in the first quadrant, y > 0, so y = \dfrac{3}{5}\sqrt{25 - x^2}.

[1]

(b)

The rectangle has width 2x and height 2y:

A = (2x)(2y) = 4xy = \frac{12}{5}\,x\sqrt{25 - x^2}

[3]

(c)

Product rule and chain rule:

\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{12}{5}\left[\sqrt{25 - x^2} + x\cdot\frac{-x}{\sqrt{25 - x^2}}\right] = \frac{12}{5}\cdot\frac{(25 - x^2) - x^2}{\sqrt{25 - x^2}}

\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{12\left(25 - 2x^2\right)}{5\sqrt{25 - x^2}}

[4]

(d)

For a stationary point, \dfrac{\mathrm{d}A}{\mathrm{d}x} = 0:

25 - 2x^2 = 0 \implies x = \frac{5}{\sqrt{2}} \quad (x > 0)

The sign of \dfrac{\mathrm{d}A}{\mathrm{d}x} changes from positive (for x slightly below \frac{5}{\sqrt2}) to negative (slightly above), so this is a maximum (and A \to 0 as x \to 0 or x \to 5).

Then y = \dfrac{3}{5}\sqrt{25 - \dfrac{25}{2}} = \dfrac{3}{5}\cdot\dfrac{5}{\sqrt{2}} = \dfrac{3}{\sqrt{2}}.

Perimeter = 2(2x + 2y) = 4(x + y) = 4\left(\dfrac{5}{\sqrt{2}} + \dfrac{3}{\sqrt{2}}\right) = \dfrac{32}{\sqrt{2}}

Perimeter = 16\sqrt{2} (the maximum area is 30)

[6]

Final answer

(b) A=\frac{12}{5}x\sqrt{25-x^2}
(d) perimeter 16\sqrt2 (largest area 30)

Question 24

[7 marks]

The curve C has equation ye^{x}+xy^{2}=2.

(a) Show that \dfrac{dy}{dx}=-\dfrac{ye^{x}+y^{2}}{e^{x}+2xy}.

[3]

(b) The normal to C at the point (0,2) meets the x-axis at N. Find the coordinates of N.

[4]

Show solutionHide solution

(a) Differentiate both sides with respect to x, using the product rule on ye^{x} and on xy^{2}: \dfrac{dy}{dx}e^{x}+ye^{x}+y^{2}+2xy\dfrac{dy}{dx}=0.

M1A1

Collect: \dfrac{dy}{dx}\left(e^{x}+2xy\right)=-\left(ye^{x}+y^{2}\right), so \dfrac{dy}{dx}=-\dfrac{ye^{x}+y^{2}}{e^{x}+2xy}.

(b) At (0,2): \dfrac{dy}{dx}=-\dfrac{2+4}{1+0}=-6.

The normal has gradient \dfrac16.

Normal: y=\dfrac16x+2.

It meets the x-axis where y=0: x=-12, so N=(-12,0).

Final answer

(b) N=(-12,0)

Question 25

[8 marks]

(a) Find \displaystyle\lim_{x\to\pi/2}\dfrac{1-\sin x}{(\pi-2x)^2}.

[5]

(b) By first writing the product as a quotient, find \displaystyle\lim_{x\to0^+}x\ln(3x).

[3]

Show solutionHide solution

(a) At x=\frac{\pi}{2} the numerator and the denominator are both 0: the form is \frac00.

Differentiate top and bottom separately: \displaystyle\lim\dfrac{-\cos x}{-4(\pi-2x)}.

M1A1

This is again \frac00, so differentiate once more: \displaystyle\lim\dfrac{\sin x}{8}.

The limit is \dfrac18.

(b) Write x\ln(3x)=\dfrac{\ln(3x)}{1/x}, which has the form \frac{-\infty}{\infty} as x\to0^+.

Differentiate top and bottom: \dfrac{1/x}{-1/x^{2}}=-x.

As x\to0^+ this tends to 0.

Final answer

(a) \frac18
(b) 0

Question 26

[18 marks]

Let f(x)=(x^{2}+a)e^{-x}, x\in\mathbb{R}, where a is a constant.

(a) Show that f'(x)=-(x^{2}-2x+a)e^{-x}.

[3]

(b) Determine the condition on a that gives the graph of f exactly two stationary points.

[3]

For the rest of this question let a=-3.

(c) Find the exact coordinates of the stationary points and determine their nature.

[5]

(d) Show that f''(x)=(x^{2}-4x-1)e^{-x}, and hence find the exact x-coordinates of the points of inflexion.

[4]

(e) The tangent to the graph at x=0 meets the x-axis at T. Find the coordinates of T.

[3]

Show solutionHide solution

(a) Product rule: f'(x)=2xe^{-x}-(x^{2}+a)e^{-x}.

M1A1

=-(x^{2}-2x+a)e^{-x}.

(b) Since e^{-x}>0, stationary points satisfy x^{2}-2x+a=0. Two distinct roots need a positive discriminant: 4-4a>0.

M1A1

So a<1.

(c) With a=-3: f'(x)=-(x-3)(x+1)e^{-x}.

f(-1)=-2e and f(3)=6e^{-3}, so the points are (-1,-2e) and (3,6e^{-3}).

A1A1

The sign of f' is -,+,- across x=-1 and x=3.

So (-1,-2e) is a minimum and (3,6e^{-3}) is a maximum.

(d) f''(x)=-(2x-2)e^{-x}+(x^{2}-2x-3)e^{-x}.

M1A1

=(x^{2}-4x-1)e^{-x}.

f''(x)=0 when x^{2}-4x-1=0, so x=2\pm\sqrt5.

The quadratic changes sign at each root, so f'' changes sign and both are points of inflexion.

(e) f(0)=-3 and f'(0)=3.

Tangent: y=3x-3.

It meets the x-axis at x=1, so T=(1,0).

Final answer

(b) a<1
(c) (-1,-2e) minimum, (3,6e^{-3}) maximum
(d) x=2\pm\sqrt5
(e) T=(1,0)

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Why a dish focuses

G1 Worked example

[11 marks]

A mirror has the shape of the curve x^2+(y-5)^2=25, for 0\le y<5. Rays of light travel vertically downwards and are reflected by the mirror. When a ray travelling vertically downwards meets a mirror at a point where the tangent has gradient m, m\ne0, the reflected ray leaves along a line with gradient \dfrac{m^2-1}{2m}.

(a) By differentiating implicitly, show that \dfrac{dy}{dx}=\dfrac{x}{5-y}.

[2]

(b) The ray along x=3 meets the mirror at P(3,1), and the ray along x=4 meets it at Q(4,2). For each ray, find the gradient of the reflected ray and the point where the reflected ray crosses the y-axis.

[4]

(c) The ray along x=h, where 0<h<5, meets the mirror at \left(h,\,5-\sqrt{25-h^2}\right). Show that the reflected ray crosses the y-axis at y=5-\dfrac{25}{2\sqrt{25-h^2}}.

[3]

(d) Find the limit of this crossing point as h\to0. Explain what (b) and (c) show about using this mirror to bring rays to a single point.

[2]

Show solutionHide solution

(a) 2x+2(y-5)\dfrac{dy}{dx}=0

\dfrac{dy}{dx}=\dfrac{-x}{y-5}=\dfrac{x}{5-y}

A1AG

(b) At P: m=\dfrac34, reflected gradient \dfrac{\frac9{16}-1}{\frac32}=-\dfrac{7}{24}

M1A1

y-1=-\dfrac7{24}(x-3); at x=0, y=1+\dfrac78=\dfrac{15}{8}

At Q: m=\dfrac43, reflected gradient \dfrac{\frac{16}9-1}{\frac83}=\dfrac{7}{24}; y-2=\dfrac7{24}(x-4) crosses at y=2-\dfrac76=\dfrac56

(c) Let c=\sqrt{25-h^2}. The point is (h,5-c) and m=\dfrac{h}{c}

reflected gradient \dfrac{\frac{h^2}{c^2}-1}{\frac{2h}{c}}=\dfrac{h^2-c^2}{2hc}

at x=0: y=5-c-h\cdot\dfrac{h^2-c^2}{2hc}=5-\dfrac{2c^2+h^2-c^2}{2c}=5-\dfrac{h^2+c^2}{2c}=5-\dfrac{25}{2\sqrt{25-h^2}}

A1AG

(d) as h\to0, \sqrt{25-h^2}\to5, so the crossing point tends to (0,\,2.5)

rays far from the axis cross lower (\frac{15}8 for h=3, \frac56 for h=4, and y decreases as h increases), so the reflected rays do not meet at one point; only a narrow beam near the axis is brought close to (0,2.5)

Final answer

(b) -\frac7{24}, (0,\frac{15}8); \frac7{24}, (0,\frac56)
(d) (0,2.5); the crossing point moves down as h grows, so there is no single focus

G1 Investigation

[12 marks]

This investigation looks at rays travelling vertically downwards that are reflected by the curve y=\dfrac{x^2}{4}. When a ray travelling vertically downwards meets a mirror at a point where the tangent has gradient m, m\ne0, the reflected ray leaves along a line with gradient \dfrac{m^2-1}{2m}.

(a) A ray travels down the line x=2. Show that its reflected ray is horizontal, and write down the point where it crosses the y-axis.

[2]

(b) Find the point where the reflected ray crosses the y-axis for the ray along x=6 and for the ray along x=-1. Make a conjecture.

[3]

(c) Prove your conjecture for the ray along x=p, where p\ne0.

[3]

(d) Find the point F through which every reflected ray passes for the curve y=kx^2, where k>0.

[2]

(e) A dish has a cross-section of the form y=kx^2. It is 60 cm across and 9 cm deep. Find how far above the bottom of the dish the point F lies.

[2]

Show solutionHide solution

(a) \dfrac{dy}{dx}=\dfrac x2=1 at x=2; reflected gradient \dfrac{1-1}{2}=0

the ray reflects along y=1, crossing the y-axis at (0,1)

(b) x=6: point (6,9), m=3, gradient \dfrac{9-1}{6}=\dfrac43, crosses at y=9-\dfrac43(6)=1

x=-1: point \left(-1,\frac14\right), m=-\frac12, gradient \dfrac{\frac14-1}{-1}=\dfrac34, crosses at y=\dfrac14+\dfrac34=1

conjecture: every reflected ray passes through (0,1)

(c) the point is \left(p,\frac{p^2}4\right) with m=\dfrac p2

reflected gradient \dfrac{\frac{p^2}4-1}{p}=\dfrac{p^2-4}{4p}

at x=0: y=\dfrac{p^2}4-p\cdot\dfrac{p^2-4}{4p}=\dfrac{p^2}{4}-\dfrac{p^2}{4}+1=1, for every p\ne0

(d) at (p,kp^2), m=2kp, reflected gradient \dfrac{4k^2p^2-1}{4kp}; at x=0, y=kp^2-\dfrac{4k^2p^2-1}{4k}

y=\dfrac1{4k}, so F=\left(0,\dfrac{1}{4k}\right)

(e) the rim is at (30,9): 9=k(30)^2, k=\dfrac1{100}

F is \dfrac{1}{4k}=25 cm above the bottom

Final answer

(a) (0,1)
(b) (0,1) both times; every reflected ray passes through (0,1)
(c) the crossing is at y=1 for every p
(d) F\left(0,\frac1{4k}\right)
(e) 25 cm

G2 Light takes the quickest path

G2 Worked example

[9 marks]

A lifeguard at L is 20 m from a straight shoreline. A swimmer at S is 30 m from the shoreline, and the point of the shoreline nearest to S is 60 m along the shore from the point nearest to L. The lifeguard runs at 6 m s^{-1} to a point P on the shoreline, x m along the shore, and then swims at 1.5 m s^{-1} straight to S.

(a) Show that the time taken, T seconds, is given by T(x)=\dfrac{\sqrt{400+x^2}}{6}+\dfrac{\sqrt{900+(60-x)^2}}{1.5}.

[2]

(b) Find the value of x that minimises T, and the least time.

[2]

(c) Find the time taken if the lifeguard heads straight for S, and the time taken if the lifeguard runs to the point of the shoreline nearest to S.

[2]

(d) Let \theta_1 and \theta_2 be the angles that LP and PS make with the line through P perpendicular to the shoreline. By considering T'(x), show that at the minimum \dfrac{\sin\theta_1}{\sin\theta_2}=4.

[3]

Show solutionHide solution

(a) LP=\sqrt{20^2+x^2} and PS=\sqrt{30^2+(60-x)^2}, by Pythagoras

time = distance \div speed for each stage: T(x)=\dfrac{\sqrt{400+x^2}}{6}+\dfrac{\sqrt{900+(60-x)^2}}{1.5}

A1AG

(b) by GDC, x=52.8

least time T=30.0 s (29.98\ldots)

(c) straight for S: the path crosses the shore where \dfrac{x}{20}=\dfrac{60-x}{30}, x=24; T(24)=36.4 s

run to the nearest point: T(60)=\dfrac{\sqrt{4000}}{6}+\dfrac{30}{1.5}=30.5 s

(d) T'(x)=\dfrac{x}{6\sqrt{400+x^2}}-\dfrac{60-x}{1.5\sqrt{900+(60-x)^2}}

\sin\theta_1=\dfrac{x}{\sqrt{400+x^2}} and \sin\theta_2=\dfrac{60-x}{\sqrt{900+(60-x)^2}}, so T'(x)=\dfrac{\sin\theta_1}{6}-\dfrac{\sin\theta_2}{1.5}

at the minimum T'(x)=0: \dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{6}{1.5}=4

A1AG

Check: at x=52.79, \sin\theta_1=0.935 and \sin\theta_2=0.234.

Final answer

(b) x=52.8, T=30.0 s
(c) 36.4 s; 30.5 s
(d) T'(x)=\frac{\sin\theta_1}{6}-\frac{\sin\theta_2}{1.5}=0

G2 Investigation

[11 marks]

The shoreline is the x-axis. A lifeguard starts at L(0,30) on the beach and must reach a swimmer at S(60,-20) in the sea, entering the water at P(x,0). The lifeguard runs k times as fast as they swim, where k\ge1. In units where the swimming speed is 1, the time taken is T(x)=\dfrac{\sqrt{900+x^2}}{k}+\sqrt{400+(60-x)^2}. Let x^* be the value of x that minimises T, and let \theta_1 and \theta_2 be the angles that LP and PS make with the line through P perpendicular to the shoreline. You may use the result that at the minimum \sin\theta_1=k\sin\theta_2.

(a) For k=1, explain why the quickest route is the straight line LS, and find x^*.

[2]

(b) Use your GDC to find x^* for k=2, k=4 and k=8. Make a conjecture about x^* as k increases.

[3]

(c) Show that 60-x^*=20\tan\theta_2, and hence show that 0<60-x^*<\dfrac{20}{\sqrt{k^2-1}} for k>1. Hence prove your conjecture.

[4]

(d) Find the least value of k for which (c) guarantees that the lifeguard enters the water within 1 m of the point nearest to the swimmer.

[2]

Show solutionHide solution

(a) for k=1, T is the total length LP+PS, which is least when L, P and S are collinear

LS crosses the shore where \dfrac{x}{30}=\dfrac{60-x}{20}, so x^*=36

(b) k=2: x^*=50.5; k=4: x^*=55.5; k=8: x^*=57.8

Note: award A1 for two correct values.

conjecture: x^* increases towards 60, so the lifeguard enters the water ever closer to the point opposite the swimmer

(c) in the right-angled triangle with hypotenuse PS, the vertical side is 20 and the horizontal side is 60-x^*, so \tan\theta_2=\dfrac{60-x^*}{20}

\sin\theta_2=\dfrac{\sin\theta_1}{k}<\dfrac1k, since \sin\theta_1<1

\theta_2 is acute and \tan^2\theta_2=\dfrac{\sin^2\theta_2}{1-\sin^2\theta_2} increases with \sin^2\theta_2<\dfrac{1}{k^2}, so \tan^2\theta_2<\dfrac{1}{k^2-1}, giving 0<\tan\theta_2<\dfrac{1}{\sqrt{k^2-1}}

0<60-x^*<\dfrac{20}{\sqrt{k^2-1}}\to0 as k\to\infty, so x^*\to60

(d) \dfrac{20}{\sqrt{k^2-1}}\le1\iff k^2\ge401

k\ge\sqrt{401}=20.0 (20.02\ldots)

Final answer

(a) x^*=36
(b) 50.5, 55.5, 57.8; x^*\to60
(c) bound \frac{20}{\sqrt{k^2-1}}\to0
(d) k=\sqrt{401}\approx20.0

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Gradient at a point	`MATH, 8:nDeriv(expression, X, value)`
Intersections	`Graph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 14 teaches differentiation from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book