Differential calculus — Mathematics Grade 12

9 exam-style question sets on Differential calculus (Paper 1), each with a hint and a fully worked answer. The app holds 45 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

First principles, differentiation rules & tangent gradientsCubic with a repeated root: turning points, sketch & tangent at the point of inflectionOptimisation: minimum area with fixed bordersFirst principles, rules of differentiation, tangents & inversesInterpreting cubic graphsRates of change (motion)First principles & rules of differentiationCubic graphs: turning points, concavity & intersecting linesOptimisation (maximum volume)

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Question 1

First principles, differentiation rules & tangent gradientsRoutine13 marksPaper 1

Show ALL your working in each part of this question.

aDetermine $f^{'} (x)$ from FIRST PRINCIPLES if $f (x) = - 5 x^{2}$ . (5)
bDetermine $\frac{d y}{d x}$ if $y = 2 x^{3} - 5 x^{2} + 7 x - 1$ . (2)
cDetermine $D_{x} [9 3 x^{2} + \frac{2}{x ^{4}}]$ . Write your answer with positive exponents. (3)
dDetermine the values of $x$ for which the tangent to the graph of $h (x) = 2 x^{3} + 3 x^{2} - 36 x + 5$ has a positive gradient. (3)

Hint

In part (a), expand $f (x + h)$ , simplify $f (x + h) - f (x)$ and take out the common factor $h$ BEFORE letting $h \to 0$ . In part (c), rewrite the cube root and the fraction as powers of $x$ first. In part (d), the gradient of the tangent at any point is $h^{'} (x)$ , so solve the quadratic inequality $h^{'} (x) > 0$ .

Worked answer

a) $f (x + h) = - 5 (x + h)^{2} = - 5 x^{2} - 10 x h - 5 h^{2}$ ✓
$f (x + h) - f (x) = - 5 x^{2} - 10 x h - 5 h^{2} + 5 x^{2} = - 10 x h - 5 h^{2}$ ✓
$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{- 10 x h - 5 h ^{2}}{h}$ ✓
$= h \to 0 lim \frac{h ( - 10 x - 5 h )}{h} = h \to 0 lim (- 10 x - 5 h)$ ✓
$= - 10 x$ ✓

b) $\frac{d y}{d x} = 6 x^{2} - 10 x + 7$ ✓✓

c) $D_{x} [9 x^{\frac{2}{3}} + 2 x^{- 4}]$ ✓ (index form)
$= 6 x^{- \frac{1}{3}} - 8 x^{- 5}$
$= \frac{6}{3 x} - \frac{8}{x ^{5}}$ ✓✓ (one ✓ per term)

d) $h^{'} (x) = 6 x^{2} + 6 x - 36$ ✓
Positive gradient: $h^{'} (x) > 0$ , i.e. $6 (x^{2} + x - 6) > 0$ , so $6 (x + 3) (x - 2) > 0$ ✓ (critical values $x = - 3$ and $x = 2$ )
$y = h^{'} (x)$ is a parabola that opens upwards, so it is positive outside its roots: $x < - 3$ or $x > 2$ ✓

Modelled on Nov 2023 Paper 1, Question 7 — same skills, new scenario.

Question 2

Cubic with a repeated root: turning points, sketch & tangent at the point of inflectionComplex18 marksPaper 1

The cubic function $f$ is defined by $f (x) = x^{3} + 9 x^{2} + 24 x + 16$ , which factorises as $f (x) = (x + 4)^{2} (x + 1)$ .

aFind the $x$ - and $y$ -coordinates of each turning point of $f$ . (4)
bSketch the graph of $f$ . Show clearly the coordinates of every intercept with the axes and of both turning points. (4)
cUse your sketch to find all values of $k$ such that the equation $f (x) = k$ has THREE distinct real roots. (2)
dThe straight line $g (x) = m x + c$ is the tangent to the graph of $f$ at its point of inflection. Determine the equation of $g$ . (6)
eCalculate the size of the acute angle between $g$ and the $x$ -axis. Round off your answer to TWO decimal places. (2)

Hint

Differentiate the expanded form, set $f^{'} (x) = 0$ and factorise — the squared factor $(x + 4)^{2}$ already tells you that one turning point sits ON the $x$ -axis, because the graph only touches the axis at a repeated root. For part (c), slide a horizontal line $y = k$ up and down your sketch and count how often it meets the graph. For the tangent, solve $f^{''} (x) = 0$ first; the gradient of the tangent is $f^{'}$ evaluated at that $x$ -value.

Worked answer

a) $f^{'} (x) = 3 x^{2} + 18 x + 24$ ✓ Set $f^{'} (x) = 0$ : $3 (x^{2} + 6 x + 8) = 3 (x + 4) (x + 2) = 0$ ✓ so $x = - 4$ or $x = - 2$ ✓ $f (- 4) = 0$ and $f (- 2) = (- 2 + 4)^{2} (- 2 + 1) = - 4$ ✓ Turning points: $(- 4; 0)$ , a local maximum ON the $x$ -axis (from the repeated factor $(x + 4)^{2}$ ), and $(- 2; - 4)$ , a local minimum.

b) A correct sketch shows:

shape: positive leading coefficient, so the graph rises from the bottom left and ends at the top right ✓
$x$ -intercepts $(- 4; 0)$ , where the graph only TOUCHES the $x$ -axis (repeated root), and $(- 1; 0)$ , where it cuts through ✓
$y$ -intercept $(0; 16)$ ✓
turning points: local maximum $(- 4; 0)$ and local minimum $(- 2; - 4)$ ✓

c) $f (x) = k$ has three distinct real roots when the horizontal line $y = k$ cuts the graph in three places, i.e. when it lies strictly between the $y$ -values of the two turning points: $- 4 < k < 0$ ✓✓ (for $k = - 4$ or $k = 0$ the line touches the graph at a turning point, so two of the roots are equal).

d) $f^{''} (x) = 6 x + 18$ ✓ $6 x + 18 = 0 \Rightarrow x = - 3$ ✓ $f (- 3) = (- 3 + 4)^{2} (- 3 + 1) = - 2$ , so the point of inflection is $(- 3; - 2)$ ✓ $m = f^{'} (- 3) = 3 (- 3)^{2} + 18 (- 3) + 24 = - 3$ ✓ $y - (- 2) = - 3 (x - (- 3))$ ✓ $g (x) = - 3 x - 11$ ✓

e) $tan θ = ∣ m ∣ = 3$ ✓ $θ = 71, 5 7^{\circ}$ ✓ (Because $m < 0$ , the inclination of $g$ itself is $18 0^{\circ} - 71, 5 7^{\circ} = 108, 4 3^{\circ}$ ; the ACUTE angle between $g$ and the $x$ -axis is $71, 5 7^{\circ}$ .)

Modelled on Nov 2023 Paper 1, Question 8 — same skills, new scenario.

Question 3

Optimisation: minimum area with fixed bordersProblem-solving6 marksPaper 1

The members of a community food garden in Soweto are planning a rectangular vegetable bed with an area of $72$ m². Gravel paths will surround the bed. The paths along the western and eastern edges of the bed are each $2$ m wide, and the paths along the northern and southern edges are each $1$ m wide. The bed and the paths together form a larger rectangle, as shown in the sketch below (NOT drawn to scale; north is at the top).

The width of the bed, measured from west to east, is $x$ metres.

aShow that the total area, in m², covered by the bed and the paths together is given by $A (x) = \frac{288}{x} + 2 x + 80$ . (3)
bCalculate the width of the bed for which the total area of the bed and the paths is a minimum. (3)

This question has a diagram, shown in the app.

Hint

Use the area of the bed to write its north–south length in terms of $x$ . Each outer side of the large rectangle equals a side of the bed PLUS a path at BOTH ends of that side. For the minimum, solve $A^{'} (x) = 0$ and reject the negative root.

Worked answer

a) North–south length of the bed $= \frac{72}{x}$ m ✓
The large rectangle measures $(x + 2 + 2) = (x + 4)$ m by $(\frac{72}{x} + 1 + 1) = (\frac{72}{x} + 2)$ m, so $A (x) = (x + 4) (\frac{72}{x} + 2)$ ✓
$= 72 + 2 x + \frac{288}{x} + 8 = \frac{288}{x} + 2 x + 80$ ✓

b) $A (x) = 288 x^{- 1} + 2 x + 80$ , so $A^{'} (x) = - \frac{288}{x ^{2}} + 2$ ✓
$A^{'} (x) = 0$ : $\frac{288}{x ^{2}} = 2$ , so $x^{2} = 144$ ✓
$x = 12$ (reject $x = - 12$ , since a width is positive) ✓
The bed should be $12$ m wide (and $\frac{72}{12} = 6$ m long). ( $A^{''} (x) = \frac{576}{x ^{3}} > 0$ for $x > 0$ , so this gives a minimum.)

Modelled on Nov 2023 Paper 1, Question 9 — same skills, new scenario.

Question 4

First principles, rules of differentiation, tangents & inversesRoutine18 marksPaper 1

Answer ALL the parts below. Show ALL calculations.

For parts (d) to (f), the function $g (x) = 3 x^{2}$ is given.

aDetermine $f^{'} (x)$ if $f (x) = 4 x^{3} - 7 x$ . (2)
bDetermine $\frac{d y}{d x}$ if $y = \frac{5}{x ^{3}} + 4 x^{5}$ . (4)
cDetermine the equation of the tangent to $k (x) = x^{3} - 4 x^{2} + 3 x + 5$ at the point where $x = 2$ . (3)
dDetermine $g^{'} (x)$ from FIRST PRINCIPLES. (5)
eWrite down ONE way in which the domain of $g$ can be restricted so that $g^{- 1}$ is a function. (1)
fThe domain of $g$ is now restricted to $x \leq 0$ . Determine the equation of $g^{- 1}$ in the form $y = \dots$ (3)

Hint

Rewrite every term as a power of $x$ before differentiating; the gradient of a tangent equals the derivative's value at the point of contact.

Worked answer

a) $f^{'} (x) = 12 x^{2} - 7$ ✓✓

b) $y = 5 x^{- 3} + 4 x^{\frac{5}{2}}$ ✓✓ (one ✓ per rewritten term)
$\frac{d y}{d x} = - 15 x^{- 4} + 10 x^{\frac{3}{2}}$ ✓✓

c) $k (2) = 8 - 16 + 6 + 5 = 3$ , so the point of contact is $(2; 3)$ ✓
$k^{'} (x) = 3 x^{2} - 8 x + 3$ , so $m = k^{'} (2) = 12 - 16 + 3 = - 1$ ✓
$y - 3 = - 1 (x - 2)$ , giving $y = - x + 5$ ✓

d) $g^{'} (x) = h \to 0 lim \frac{g ( x + h ) - g ( x )}{h}$ ✓
$= h \to 0 lim \frac{3 ( x + h ) ^{2} - 3 x ^{2}}{h}$ ✓
$= h \to 0 lim \frac{3 x ^{2} + 6 x h + 3 h ^{2} - 3 x ^{2}}{h}$ ✓
$= h \to 0 lim \frac{h ( 6 x + 3 h )}{h} = h \to 0 lim (6 x + 3 h)$ ✓
$= 6 x$ ✓

e) $x \geq 0$ ✓ (OR: $x \leq 0$ )

f) Swap $x$ and $y$ : $x = 3 y^{2}$ ✓ so $y^{2} = \frac{x}{3}$ and $y = \pm \frac{x}{3}$ ✓
The restricted domain $x \leq 0$ means the range of $g^{- 1}$ is $y \leq 0$ , so $y = - \frac{x}{3}, x \geq 0$ ✓

Modelled on Nov 2024 Paper 1, Question 8 — same skills, new scenario.

Question 5

Interpreting cubic graphsRoutine8 marksPaper 1

The sketch below shows the graph of a cubic function $f$ with a positive leading coefficient. $A (1; 18)$ is the local maximum turning point and $B (5; - 14)$ is the local minimum turning point of $f$ . The graph cuts the $y$ -axis at $C (0; 11)$ . The equation of $f$ is NOT given.

aWrite down the values of $x$ for which $f$ is strictly decreasing. (2)
bWrite down the $x$ -intercepts of the graph of $y = f^{'} (x)$ . (2)
cFor which values of $x$ is the graph of $f$ concave up? (2)
dDetermine the values of $k$ for which $f (x) + k$ will have THREE positive $x$ -intercepts. (2)

This question has a diagram, shown in the app.

Hint

Every answer can be read from the two turning points and the y-intercept — no equation is needed; for the last part, think about what adding k does to each of those three y-values.

Worked answer

a) $f$ decreases where the graph falls, i.e. between the turning points: $1 < x < 5$ ✓✓

b) $f^{'} (x) = 0$ exactly at the turning points of $f$ , so $y = f^{'} (x)$ cuts the $x$ -axis at $x = 1$ ✓ and $x = 5$ ✓

c) The point of inflection of a cubic lies midway between its turning points: $x = \frac{1 + 5}{2} = 3$ ✓, and since the graph rises again after $B$ , $f$ is concave up for $x > 3$ ✓

d) Adding $k$ shifts every point vertically by $k$ . For THREE $x$ -intercepts the new maximum must lie above the axis: $18 + k > 0$ , so $k > - 18$ ✓. For all three intercepts to be POSITIVE the new $y$ -intercept must still be below the axis: $11 + k < 0$ , so $k < - 11$ (then the minimum $- 14 + k$ is also below the axis). ✓ Therefore $- 18 < k < - 11$

Modelled on Nov 2024 Paper 1, Question 9 — same skills, new scenario.

Question 6

Rates of change (motion)Problem-solving8 marksPaper 1

A racing drone accelerates from rest along a straight course and then slows down again. Its speed after $t$ seconds is given by $v (t) = - 3 t^{2} + 18 t$ metres per second, where $v (t) = s^{'} (t)$ and $s (t)$ is the distance, in metres, that the drone has covered after $t$ seconds.

aCalculate the maximum speed that the drone reaches. (3)
bCalculate the TOTAL distance that the drone covers from the start until the moment it comes to rest again. (5)

Hint

Speed peaks where its own derivative is zero; for the distance, first recover s(t) from s'(t) using s(0) = 0, then substitute the time at which the speed returns to zero.

Worked answer

a) $v^{'} (t) = - 6 t + 18$ ✓. Maximum speed where $v^{'} (t) = 0$ : $t = 3$ s ✓. $v (3) = - 3 (9) + 18 (3) = 27$ m/s ✓

b) $s^{'} (t) = - 3 t^{2} + 18 t$ , so $s (t) = - t^{3} + 9 t^{2} + c$ ✓✓ and $s (0) = 0$ gives $c = 0$ ✓. The drone is at rest when $v (t) = 0$ : $- 3 t (t - 6) = 0$ , so $t = 6$ s (since $t > 0$ ) ✓. Total distance $= s (6) = - (216) + 9 (36) = 108$ m ✓

Modelled on Nov 2024 Paper 1, Question 10 — same skills, new scenario.

Question 7

First principles & rules of differentiationRoutine10 marksPaper 1

Answer ALL the parts below. Show ALL calculations.

aDetermine $f^{'} (x)$ from FIRST PRINCIPLES if $f (x) = 3 x - 5$ . (4)
bDetermine $g^{'} (x)$ if $g (x) = 2 x^{4} + 3 x^{2}$ . (2)
cDetermine $\frac{d y}{d x}$ if $y = \frac{2 x ^{5} - 3}{x ^{2}}$ . (4)

Hint

Substitute $x + h$ into $f$ and simplify the numerator until $h$ cancels; in part (c) split the fraction into two separate powers of $x$ before differentiating.

Worked answer

a) $f (x + h) = 3 (x + h) - 5 = 3 x + 3 h - 5$ ✓
$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{( 3 x + 3 h - 5 ) - ( 3 x - 5 )}{h}$ ✓
$= h \to 0 lim \frac{3 h}{h} = h \to 0 lim 3$ ✓
$= 3$ ✓

b) $g^{'} (x) = 8 x^{3} + 6 x$ ✓✓

c) $y = \frac{2 x ^{5}}{x ^{2}} - \frac{3}{x ^{2}} = 2 x^{3} - 3 x^{- 2}$ ✓✓ (one ✓ per rewritten term)
$\frac{d y}{d x} = 6 x^{2} + 6 x^{- 3} = 6 x^{2} + \frac{6}{x ^{3}}$ ✓✓

Modelled on Nov 2025 Paper 1, Question 8 — same skills, new scenario.

Question 8

Cubic graphs: turning points, concavity & intersecting linesProblem-solving17 marksPaper 1

The sketch below shows the graph of $f (x) = x^{3} - 3 x^{2} - 9 x + 2$ . The graph cuts the $x$ -axis at A, B and C. D is the local maximum turning point and E is the local minimum turning point of $f$ . The sketch is NOT drawn to scale.

aDetermine the coordinates of E, the minimum turning point of $f$ . (4)
bFor which values of $x$ is the graph of $f$ concave down? (3)
cIt is further given that $A (- 2; 0)$ . Determine the values of $x$ for which $f (x) \cdot f^{''} (x) < 0$ . (4)
dThe line $y = - 9 x + t$ , where $t$ is a constant, cuts the graph of $f$ at THREE distinct points. Determine the values of $t$ . (6)

This question has a diagram, shown in the app.

Hint

Turning points come from $f^{'} (x) = 0$ and concavity from the sign of $f^{''} (x)$ ; for the last part, find the two tangents to $f$ that are parallel to the given line.

Worked answer

a) $f^{'} (x) = 3 x^{2} - 6 x - 9$ ✓
$3 (x^{2} - 2 x - 3) = 3 (x - 3) (x + 1) = 0$ , so $x = 3$ or $x = - 1$ ✓
$f^{''} (x) = 6 x - 6$ and $f^{''} (3) = 12 > 0$ , so the minimum is at $x = 3$ ✓
$f (3) = 27 - 27 - 27 + 2 = - 25$ , so $E (3; - 25)$ ✓

b) $f^{''} (x) = 6 x - 6$ ✓
Concave down where $f^{''} (x) < 0$ , i.e. $6 x - 6 < 0$ ✓
$x < 1$ ✓

c) $f (- 2) = - 8 - 12 + 18 + 2 = 0$ , so $(x + 2)$ is a factor and $f (x) = (x + 2) (x^{2} - 5 x + 1)$ ✓
$x^{2} - 5 x + 1 = 0$ gives $x = \frac{5 - 21}{2} \approx 0, 21$ at B and $x = \frac{5 + 21}{2} \approx 4, 79$ at C ✓
Opposite signs are needed. $f > 0$ between A and B, and there $f^{''} < 0$ (since $x < 1$ ): $- 2 < x < \frac{5 - 21}{2}$ ✓
$f < 0$ between B and C, but $f^{''} > 0$ only for $x > 1$ : $1 < x < \frac{5 + 21}{2}$ ✓

d) The line cuts the cubic three times only if it lies strictly between the TWO tangents to $f$ with gradient $- 9$ ✓
$f^{'} (x) = - 9$ : $3 x^{2} - 6 x - 9 = - 9$ , so $3 x (x - 2) = 0$ and $x = 0$ or $x = 2$ ✓
At $x = 0$ : $f (0) = 2$ , so the tangent is $y = - 9 x + 2$ ✓
At $x = 2$ : $f (2) = - 20$ and $- 20 = - 9 (2) + c$ gives $c = - 2$ , so the tangent is $y = - 9 x - 2$ ✓
For THREE distinct points of intersection, $t$ must lie strictly between the two $y$ -cuts ✓
$- 2 < t < 2$ ✓

Modelled on Nov 2025 Paper 1, Question 9 — same skills, new scenario.

Question 9

Optimisation (maximum volume)Problem-solving6 marksPaper 1

A plumber makes a rainwater drum from a rectangular steel sheet $A B C D$ with $A B = x$ cm, $B C = h$ cm and a perimeter of $480$ cm. The sheet is curled round, without any overlap, so that side $A B$ bends into a complete circle: $A B$ becomes the circumference of the base circle of a cylinder and $B C$ becomes its height. A circular steel disc is welded onto the bottom edge, and the top of the drum stays open.

aShow that the volume of the drum, in cubic centimetres, is given by $V (x) = \frac{240 x ^{2} - x ^{3}}{4 π}$ . (3)
bCalculate the value of $x$ for which the drum will have a maximum volume. (3)

Hint

Use the perimeter to write $h$ in terms of $x$ and the curled side to get $r = \frac{x}{2 π}$ ; substitute into $V = π r^{2} h$ , then set $V^{'} (x) = 0$ , factor out $x$ and reject the root that forms no drum.

Worked answer

a) Perimeter: $2 (x + h) = 480$ , so $h = 240 - x$ ✓
$A B$ forms the circumference: $2 π r = x$ , so $r = \frac{x}{2 π}$ ✓
$V = π r^{2} h = π (\frac{x}{2 π})^{2} (240 - x) = \frac{x ^{2} ( 240 - x )}{4 π} = \frac{240 x ^{2} - x ^{3}}{4 π}$ ✓

b) $V^{'} (x) = \frac{480 x - 3 x ^{2}}{4 π}$ ✓
$V^{'} (x) = 0$ gives $3 x (160 - x) = 0$ , so $x = 0$ or $x = 160$ ✓
Reject $x = 0$ , since no drum is formed ( $0 < x < 240$ ); the volume is a maximum at $x = 160$ ✓

Modelled on Nov 2025 Paper 1, Question 10 — same skills, new scenario.

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