KT Classroom Unit 9: Differentiation

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

Preparation for the Mathematics examination brought to you by Kagiso Trust

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

Preparation for the Mathematics examination brought to you by Kagiso Trust

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