Differentiation: The Rules and How Exams Use Them — Worksheet & Memo

∑ DANEMATHICS

FREE CAPS MATHS RESOURCES

Grade 12 · Calculus

Differentiation: The Rules and How Exams Use Them

MARKING GUIDELINE

Marks

Duration

55 minutes

Questions

Name:

Class:

Date:

Mark

/ 35

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
If y = 4x⁵ − 6x⁴ + 3x, then dydx = …
(1)
A)20x⁴ − 24x³ + 3✓
B)20x⁴ − 24x³ + 3x
C)20x⁵ − 24x⁴ + 3
D)4x⁴ − 6x³ + 3
Answer: A — Multiply by the power, then reduce the power by one.
- B — the derivative of 3x is 3, not 3x
- C — the powers must DROP by one, not stay the same
- D — the coefficients must be multiplied by the old power
1.2
The derivative of a CONSTANT is …
(1)
A)0✓
B)1
C)the constant itself
D)undefined
Answer: A — A constant function has a horizontal graph, whose gradient is zero.
- B — that is the derivative of x
- C — a constant does not change, so its rate of change is zero
- D — constants are perfectly differentiable
1.3
If f(x) = √x, then f'(x) = …
(1)
A)12√x✓
B)1√x
C)2√x
D)12x
Answer: A — Write √x as x^½; its derivative is ½x^−½ = 1 ÷ (2√x).
- B — forgot the factor of ½ that the power brings down
- C — differentiated as though the power were increasing
- D — reduced the power by one but dropped the square root
1.4
If y = 3x, then dydx = …
(1)
A)−3x²✓
B)3x²
C)−3x
D)−3x
Answer: A — Write it as 3x⁻¹; the derivative is −3x⁻².
- B — the power of −1 brings down a MINUS
- C — the power must drop from −1 to −2
- D — the power drops; it does not become +1
1.5
Before differentiating (3x + 2)(x − 1), you should …
(1)
A)expand the brackets first✓
B)differentiate each bracket and multiply the results
C)differentiate the first bracket only
D)take logarithms
Answer: A — The Grade 12 syllabus has no product rule, so the expression is expanded into a sum of powers first.
- B — differentiating each factor and multiplying does NOT give the derivative of a product
- C — both factors depend on x, so neither may be left alone
- D — logarithms are for exponential equations, not for this
1.6
If f(x) = x³ − 3x, then f'(2) = …
(1)
A)9✓
B)2
C)12
D)6
Answer: A — f'(x) = 3x² − 3, so f'(2) = 12 − 3 = 9.
- B — substituted into f(x) instead of f'(x): 8 − 6 = 2
- C — forgot to subtract the 3
- D — differentiated 3x to 3x² instead of to 3
1.7
D_x notation means …
(1)
A)differentiate with respect to x✓
B)the value of the function at x
C)the domain of the function
D)the discriminant
Answer: A — D_x, dy/dx and f'(x) all mean the same operation.
- B — that is f(x)
- C — the domain is the set of allowed inputs, not an operation
- D — the discriminant belongs to a quadratic equation
1.8
The gradient of the TANGENT to y = x² at x = 3 is …
(1)
A)6✓
B)9
C)3
D)2
Answer: A — dy/dx = 2x, so at x = 3 the gradient is 6.
- B — that is the y-VALUE at x = 3, not the gradient
- C — that is the x-value itself
- D — that is the coefficient in the derivative, before substituting
1.9
If f'(a) = 0, the graph of f has …
(1)
A)a stationary point at x = a✓
B)an x-intercept at x = a
C)a point of inflection at x = a
D)an asymptote at x = a
Answer: A — A zero gradient means the tangent there is horizontal.
- B — an x-intercept needs f(a) = 0, not f'(a) = 0
- C — a point of inflection needs f''(a) = 0, which is a different condition
- D — an asymptote is where the function is undefined, not where its gradient is zero
1.10
If y = (2x)³, then dydx = …
(1)
A)24x²✓
B)6x²
C)8x²
D)3(2x)²
Answer: A — (2x)³ = 8x³, whose derivative is 24x².
- B — cubed only the x, leaving the 2 outside
- C — differentiated the 8 but not the x³
- D — left the bracket unexpanded, which the Grade 12 rules do not allow

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[13 MARKS]

Answer the questions below. Show ALL your working.

2.1
Determine f′(x) from FIRST PRINCIPLES if f(x) = 2x² − 3x.
(5)
f′(x) = lim_h→0 f(x + h) − f(x)h (1)
= lim_h→0 2(x + h)² − 3(x + h) − (2x² − 3x)h (1)
= lim_h→0 4xh + 2h² − 3hh (1)
= lim_h→0 (4x + 2h − 3) (1)
= 4x − 3 (1)
2.2
Determine dydx if y = 4x⁵ − 6x⁴ + 3x.
(3)
dydx = 20x⁴ − 24x³ + 3 (3)
2.3
Determine D_x [ (3x2 − 13x)² ].
(5)
Expand FIRST — you may not differentiate a bracket term by term (1)
= 9x²4 − 1 + 19x² (2)
Write the last term as 19x⁻² (1)
= 9x2 − 29x³ (1)

Question 3

[12 MARKS]

The graph of h(x) = ax³ + bx² has turning points at the origin O(0 ; 0) and at B(4 ; 32). A is the other x-intercept of h.

3.1
Show that a = −1 and b = 6.
(5)
h′(x) = 3ax² + 2bx (1)
h′(4) = 0: 48a + 8b = 0 (1)
h(4) = 32: 64a + 16b = 32 (1)
From the first, b = −6a (1)
64a − 96a = 32 gives a = −1, so b = 6 (1)
3.2
Calculate the coordinates of A.
(3)
−x³ + 6x² = 0 (1)
x²(6 − x) = 0 (1)
A(6 ; 0) (1)
3.3
Write down the values of x for which h is INCREASING.
(2)
h rises between its turning points (1)
0 < x < 4 (1)
3.4
Write down the values of x for which h is CONCAVE DOWN.
(2)
h″(x) = −6x + 12 < 0 (1)
x > 2 (1)

TOTAL: 35 marks

This question paper consists of 3 questions.