MATHEMATICS Learner’s Study and Revision Guide for Grade 12 DIFFERENTIATION
Revision Notes, Exercises and Solution Hints by
Roseinnes Phahle Examination Questions by the Department of Basic Education
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Contents Unit 9 Revision notes
3
Differentiation from first principles
3
Exercise 9.1
4
Differentiation using the rule
4
Exercise 9.2
5
Answers
6
Examination questions with solution hints and answers
7
More questions from past examination papers
10
Answers
13
How to use this revision and study guide 1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic in class or from a textbook. 2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty. 3. The notes and exercises are followed by questions from past examination papers. 4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes. 5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty. 6. What follows next are more questions taken from past examination papers. 7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge. 8. Finally, don’t be a loner. Work through this guide in a team with your classmates.
Differentiation
REVISION UNIT 9: DIFFERENTIATION Recall that the symbols
D x or
dy or f ' ( x) dx
stand for the derived function of y = f (x ) after differentiation and that they represent the gradient function at any point x on the curve of y = f (x ) .
dy is read as dee y dee x or dee dee x of y. Dee dee x is an operator: it operates on a function f ( x ) to dx produce a gradient function f ' ( x ) . It does not imply multiplication or division. Recall also that the gradient to a point on a curve is defined by the gradient of the tangent to the curve at that point. It is crucial to understand this as you will see on page 67. The following are two ways of differentiating. Differentiation rule
Differentiation from first principles
If y = f ( x ) = ax n then
D x or
For any expression y = f (x ) we have
dy or f ' ( x ) = nax n −1 dx
D x or
dy f ( x + h) − f ( x) or f ' ( x ) = Limit h → 0 dx h
Given y = f (x ) ,you will be required to differentiate from first principles or using a rule.
DIFFERENTIATION FROM FIRST PRINCIPLES dy f (x + h ) − f (x ) where f ( x + h ) is simply f ( x ) with x replaced by x + h . = Lim h → 0 dx h This formula gives you a recipe: Step 1: determine f ( x + h ) by replacing x in the function f (x ) by x + h . Step 2: subtract f (x ) from f ( x + h ) and simplify as far as possible, extracting common factors if there are any. Step 3: divide [ f ( x + h) − f ( x ) ] by h and cancel out common factor in numerator and denominator. Step 4: find the limit as h tends to zero. 3
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Example of differentiation from first principles Find the gradient of the curve f ( x) = 3 x 2 + 4 x − 7 at any point where x . Solution – the four steps indicated above are now illustrated by applying to this example.
Given f ( x ) = 3x 2 + 4 x − 7
Step 3:
Step 1: f ( x + h) = 3( x + h) 2 + 4( x + h) − 7 Step 2: f ( x + h) − f ( x )
f ( x + h) − f ( x ) h h(6 x + 3h + 4) = h = 6 x + h + 4
= {3( x + h) + 4( x + h) − 7} − {3 x + 4 x − 7}
= {3( x 2 + 2 xh + h 2 ) + 4 x + 4h − 7} − 3 x 2 − 4 x + 7 Step 4: Limit
2
2
f ( x + h) − f ( x ) h = Limit (6 x + h + 4) h →0
= 3 x 2 + 6 xh + 3h 2 + 4 x + 4h − 7 − 3 x 2 − 4 x + 7 h→0 = 6 xh + 3h 2 + 4h = 6 x + 0 + 4 = h (6 x + 3h + 4) 6x + 4 = Thus the gradient at any point on the curve of f ( x) = 3 x 2 + 4 x − 7 is given by 6 x + 4 .
EXERCISE 9.1 Differentiate the following from first principles: 1.
f (x ) = 4 x 2 − 5 x + 3
3. f ( x ) =
2. f ( x ) =
1 x
1 4. f ( x ) = 5x 3 2 x
DIFFERENTIATION USING THE RULE If y = ax n then
dy = nax n −1 dx
This means that each term must be in the form ax n before you can differentiate. This further means that rational expressions (that is, in the form of a fraction) must be divided out in order to get each term in the form ax n . For example, y = 2
2 5 x2 3
−2
must be transformed to the form 5
−1 dy 2 2 4 − 4 1 2 − 2 2 =− × x3 =− x 3 =− . . y = x 3 where a = and n = − . Then dx 3 5 15 15 3 x 5 5 5 3
Differentiation
NOTE: When differentiating from first principles always replace y by f ( x ) .
EXERCISE 9.2 1.
Differentiate each of the following from first principles:
1.1
y = 4x
1.2
y = 4x 2
1.3
y = 4x 3
1.4
y = 4x + 3
1.5
3x 2 + 5 x − 7
1.6
y = x 3 − 4x
1.7
y=
1.8
y=
1 x2
1.9
y = x+
1.10
y=
3 x2
2.
Use the rule to differentiate each of the following:
2.1
y = 3x 7
1 x 1 x
2.2
y=
x
2.4
y=
4 x2
2.6
y=
1
2.8
y=
2.9
y=(2x‐1)(x+3)
2.10
y=
3.
Prove that D x ⎢3 x 4 −
4.
Prove that D x ⎢
2.3
y=
2 x
2.5
y = 3 x2
x
2.7
y=
x2 − 3 x2
x+2 x
⎡ ⎣
1 ⎤ 43 1 = x+ 3 2 ⎥ 2x ⎦ 3 x
⎡ x 1⎤ 1 3 − 3⎥= + 4 x ⎦ 4 x x ⎣ 2
5
3x 3 + 5x 2 − 6 x2
Preparation for the Mathematics examination brought to you by Kagiso Trust
ANSWERS EXERCISE 9.1 NOTE: You are asked to differentiate from first principles so you will only get credit for doing just that. Always express your answers with positive indices.
dy = 8x − 5 dx dy 1 2. = x − 2 = 2 dx x dy 2 3. = −2 x −3 = − 3 dx x dy 5. = 5x 2 dx EXERCISE 9.2
1.
NOTE: You must have used first principles to obtain the following answers:
dy = 4 dx dy 1.2 = 8x dx dy 1.3 = 12x 2 dx dy 1.4 = 4 dx dy 1.5 = 6x + 5 dx dy 2 1.6 = 3x − 4 x dx dy 1 1.7 = − x −2 = − 2 dx x dy 2 1.8 = −2 x −3 = − 3 dx x dy 1 1.9 = 1− 2 dx x dy 6 1.10 =− 3 dx x
1.1
See if you can convert your answers to look like the ones below by expressing your answers with positive indices and where you have fractional
indices by expressing the answers with radical signs.
dy dx dy 2.2 dx dy 2.3 dx dy 2.4 dx
2.1
= 21x 6
=
1
2 x 2 =− 2 x 8 =− 3 x
dy 2 = 3 dx 3 x dy 1 =− 2.6 dx 2 x3 2.5
In cases where you are differentiating rational
ax r + bx s you expressions such as, for example, cx t must convert the expression as follows:
ax r bx s + cx t cx t
and then simply each term by dividing out the x ’s.
Only then will you be able to apply the rule to each term.
dy 6 = 3 dx x dy 1 1 2.8 = −3 dx 2 x x dy 2.9 = 4x + 5 dx dy 12 2.10 = 3+ 3 dx x
2.7
Differentiation
PAPER 1 QUESTION 10
DoE/ADDITIONAL EXEMPLAR 2008
PAPER 1 QUESTION 8
DoE/NOVEMBER 2008
PAPER 1 QUESTION 9
DoE/PREPARATORY EXAMINATION 2008
7
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PAPER 1 QUESTION 10
Number Hints and answers To differentiate from 1st principles, substitute 10.1
Work out the solutions in the boxes below
into:
f ' ( x ) = Lim h →0
f (x + h ) − f (x ) h
This formula gives you a recipe:
Step 1: determine f ( x + h ) by replacing x in the function f (x ) by x + h .
Step 2: take f (x ) away from f ( x + h ) and simplify as far as possible, extracting common factors if any. Step 3: divide [ f ( x + h) − f ( x ) ] by h and cancel out common factor in numerator and denominator. Step 4: find the limit as h tends to zero.
Answer: f ' ( x ) = 3x The differentiation rule is: 2
10.2.1
If y = ax n then
dy = nax n −1 dx
which requires each term to be written in the n
form ax in order to carry out its differentiation.
Answer:
dy 1 1 =− − dx 5 x 3 33 x 2 10.2.2
n
Divide to get each term in the form ax That is write as:
x 4 3x 2 7 − + y = x x x n
Simplify, so every term is of the form ax Answer:
dy 7 = 3x 2 − 3 − 2 dx x
DoE/ADDITIONAL EXEMPLAR 2008
Differentiation
PAPER 1 QUESTION 8
Number Hints and answers 8.1 Follow the steps given in 10.1 above. Answer: f ' (x) − 6 x 8.2 Follow the hints given above. Answer:
DoE/NOVEMBER 2008
Work out the solutions in the boxes below
dy 1 1 = + 4 dx 4 x 2 x PAPER 1 QUESTION 9
DoE/PREPARATORY EXAMINATION 2008
Number Hints and answers 9.1.1 Follow the steps given in 10.1 above. Answer: f ' (x) − 2 x 9.1.2 Find f (1) .
9.2.1
Find f (3) . Then determine the average gradients using the formula for the gradient of a line joining two given points. Answer: Average gradient = ‐4 Multiply out the brackets, collect like terms and differentiate. Answer:
9.2.2
Work out the solutions in the boxes below
dy = 5x 4 − 6 x 2 + 2 x dx
Divide as in 10.2.2 above and differentiate.
3
− dy Answer: = 2x 2 dx
9
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MORE QUESTIONS FROM PAST EXAMINATION PAPERS Exemplar 2008
Preparatory Examination 2008
Differentiation
Feb – March 2009
November 2009 (Unused paper)
11
Preparation for the Mathematics examination brought to you by Kagiso Trust
November 2009 (1)
Feb – March 2010
Differentiation
ANSWERS Exemplar 2008 1 9.1 f ' ( x ) = − 2 x 9.2.1 D x − 5 x 2 + 2 x = −10 x + 2
[
9.2.2
]
1 2
dy 3 = x − x −4 dx 2
Preparatory Examination 2008 9.1.1 f ' ( x ) = − x 2 9.1.2 Average gradient = ‐4 dy 9.2.1 = 5x 4 − 6x 2 + 2x dx 3 − dy 9.2.2 = 2x 2 dx Feb/March 2009 11.1 f ' ( x ) = 2 x − 2
[
November 2009 (Unused paper) 9.1 f ' ( x ) = −10 x 9.2 D x [( x − 2 )( x + 3)] = 2 x + 1 9.3.1 Depth after 3 days = 6,75m 9.3.2 Rate of decrease in depth = 2,25m/day November 2009(1) 10.1 f ' ( x ) = −4 x 3 dy 10.2 = 2x + 4 dx 2x Feb/March 2010 1 10.1 f ' (x ) = − 2 x dy 10.2 = −20 + 50 x dx
]
11.2.1 D x (x 3 − 3) = 6 x 5 − 18 x 3 2
3
11.2.1
− dy x2 = −2 x 2 − dx 3
13