introduction to differentiation i basic functions f servello

Introduction to

Differentiation IBasic Functions

F Servello

Table of contents

(1) Differentiating Polynomial Expressions (Basic): axn

GO!

(2) The Chain Rule: (f(x))n GO!

(3) The Product Rule: f(x).g(x) GO!

(4) The Quotient Rule: f(x)/g(x) GO!

Section 1

Background info slide #1

Differentiating, or finding the derivative, means finding the instantaneous rate of change i.e. gradient of the tangent of a function y = f(x). It is important to know thoroughly the notations associated with this operation. Study this closely!

If the Original function is called…..

Then the Derived function should be called…..

y y’ ordx

dy

f(x) f ’ or f ’ (x)

Background info slide #2For the purposes of Section 1 of this powerpoint, we will define a POLYNOMIAL* as any expression containing terms only of the format axn, where a and n are constants and x (or any other letter) is the variable. It may be a single term, or a string of terms connected with addition and/or subtraction signs.

Examples of “polynomials” (for our purposes*) include:

3x4 5t – 7 5x – 3 + ½ x x

4x6 + 2x 4x1/2 – π 4w4 – 2w– ¼ 8Examples of non-polynomials (for our purposes*) include:

(4x – 7)8 3 sin x 4x

5x2 – 3 tan2x

3

22 x

43 x*NOTE! The definition of “Polynomial” has been adjusted for convenience here. The strict definition of “Polynomial” excludes terms with powers that are negative and/or integers. Also, factored expressions such as (4x – 7)8 can be expanded to make expressions that

are, by definition, polynomials.

Section 1. Differentiating Polynomials: axn

This is the most basic form of differentiation. The rules you must learn are as follows:

1.1 1)( nn naxaxdx

d

1.2 aaxdx

d)(

1.3 0)( constanydx

d

This is just a special case of 1.1 with n = 1. So 3x becomes 3; – 2x becomes – 2 etc…

Another special case of 1.1 with n = 0. Constants (numbers on their own) always just become 0.

x is the variable; whereas a and n are constants

Section 1: Examples to try

1 )3( 5xdx

d

2 )5( 2 xxdx

d

3 )732( 2 xxdx

d

Note x2 is treated as 1x2 so it becomes 2x1 i.e. 2x, and 5x just becomes 5.

Note the 3x becomes 3, and the – 7 disappears because it’s a constant.

Note the 3 and 5 are multiplied, and the original power (5) is lowered by 1 to give 4

4

If y = 3x2 – ½ x4 – x , find y’.

Ans: y’ = 6x – 2x3 – 1 , using previous rules 1.1, 1.2

Note – the alternative phrasing of this question.

y’ merely means dx

dy

15x4

Click to check your answers!

2x – 5

4x – 3

Many expressions don’t look like polynomials at first, but you can turn them into polynomials by removing brackets or cancelling fractions, as in the following examples. To differentiate these “disguised” polynomials successfully, you should aim to get your function looking like….

x ± x ± x ….etc

Where the squares represent numbers.

See Preliminary Notes on Slide # 4

“Disguised polynomials” : Expressions which can be put into polynomial format, i.e. axn.

Example 1.1

Given f(x) = (3x – 5)2, find f’(x).

Solution

Aim to get the (3x – 5)2 looking like a string of terms of the form axn (as on previous slide)

Expanding (3x – 5)2, we get

f(x) = 9x2 – 30x + 25 Did you remember the 30x in the middle?

(a – b)2 = a2 – 2ab + b2Which is now in polynomial format and so it’s differentiable!

Answer: f ’(x) = 18x – 30 Using our rules from before!

CAUTION! If the power were higher than a ”square”, expanding might be too onerous! There is an alternative we learn later (the Chain Rule)

Expressions which can be put into polynomial format, i.e. axn.

Example 1.2

Solution

Aim to get the (x – 7)(x + 7) looking like a string of terms of the form axn.(See Slide 9)

Expanding (x – 7)(x + 7), we get

y = x2 – 49 Did you remember the difference of 2 squares?

(a – b)(a + b) = a2 – b2Which is now in polynomial format and hence differentiable!

Answer: y’ = 2x Using our rules from before!

If y = (x – 7)(x + 7), find dx

dy


When multiplying a whole number or letter by a fraction, the whole number multiplies into the TOP of the fraction!

5

6

5

23

7

3

7

3 xx 2

2

9

5

9

5x

x

Ability to manipulate these expressions will be very handy as we move into more difficult examples in later slides.

Example Differentiate5

3 8x

Solution Use (3) above to rewrite as5

3 8x 8

5

3x

Differentiating gives =dx

dy 7

5

24x Ans!

Note 5

24

5

38


Example 1.3

Solution

Given y = , find dx

dy

5

3x

We need to get the into the format axn.. This requires a simple manipulation from

5

3x

5

3xto x

5

3 which is now of the format axn [Note a = & n = 1]

5

3

Answer:

5

3

dx

dy


Example 1.4

Solution

Given y = find dx

dy

2

34

5

32

x

xx

Decompose the fraction as follows:

2

34

5

32

x

xx2

3

2

4

5

3

5

2

x

x

x

x

5

3

5

2 2 xx

xx5

3

5

2 2 which is now in diffrentiable format!

Now we can differentiate!

xxy5

3

5

2 2

5

3

5

4 x

dx

dyAns:


Negative and Fractional Powers can be very useful in differentiating! Learn these rules and refer to the INDICES Powerpoint for further examples

nn

xx

1

21

xx nn xx1

33

22 xx

77 3

5

3

5 xx

Make sure you can do manipulations like these…..

1

2

1

2

1 xx

21

55 xx

323

21

2222 xxxxxx

21

7

2

7

2 xx

Study this one closely!

Example 1.5 Differentiate 6

1

x

So 76 xdx

dy or 7

6

xdx

dy

Ans

6

1

xy as

6xyRewriteSolution

Example 1.6 Differentiate37

2

xSolution Rewrite 37

2

xy as

3

7

2 xy

So 4

7

6 xdx

dy or 47

6

xdx

dy

Ans

Refer to the previous slide…..

You will note that there are TWO correct answers for each example. You should practise manipulating your answers to get them out of the first format (with negative and/or fractional powers) and into the second format (without negative/fractional powers). Practising this process will greatly strengthen your understanding of the algebra required for manipulation of indices.

Example 1.7

Differentiate xy 6

xy 6 as 21

6xy RewriteSolution

So 21

3 x

dx

dyi.e.

xdx

dy 3

Ans

Note #1 Again here, the second of these two answers is the preferred one although both are correct – make sure you get plenty of practice at swapping in and out of negative & fractional power format.

Note #2 Subtracting 1 from ½ gives – ½ , which is the new power. Also 6 × ½ = 3.

Remember also that the negative in the power means that term (and that term only – in this case the x but not the 3 – is in the BOTTOM of the fraction)

Example 1.8

Differentiate 342

3)( x

xxf

31

21

42

3xxy

Rewrite asSolution

Which is now in differentiable axn format

32

23

3

4

4

3 xxdx

dy NOTE!! It may be timely here to revise how to multiply fractions. Just multiply across the top and then across the bottom. So

4

3

2

3

2

1

DID YOU KNOW? There’s a fast way to subtract 1 from any fraction (in your head) ! SHOW ME!

Ans

OR…

3 23 .3

4

4

3

xxdx

dy

AnsMake sure you can follow this!


It is helpful to be able to convert between SURDS and FRACTIONAL POWERS. For the most part the rules required stem from Year Ten work on indices. For example

xx2(a) 2

12 xx 25

x

7x(b) 217 )(x 2

7x

Add powers when multiplying: 2 + ½ = 5/2

Multiply powers when removing brackets

32

3 2

.5

2

xx

x(c)

312

312

5

)(2

xx

x

37

32

5

2

x

x 35

5

2 x

Subtract powers when dividing

Example 1.9

Differentiate 3

2

.5

23

xx

xxxy

33

2

.5

2

.5

3

xx

xx

xx

xy Split first:Solution

Change surds to powers:

31

21

31

2

.5

.2

.5

3

xx

xx

xx

xy

Simplify each top & bottom by adding powers: 3

4

23

34

2

5

2

5

3

x

x

x

xy

Extract fractions to front & subtract powers

61

32

5

2

5

3xxy

Which is now differentiable!

61

32

5

2

5

3xxy

65

31

15

1

5

2 xx

dx

dy

Remember – it’s a good idea to practise turning this index notation back into surd format

Ans

6 53 .15

1

.5

2

xxdx

dy Ans

Example 1.10

Find the equation of the tangent and normal to the parabola y = x2 – 2x at the point where x = – 1.

It may be useful to know what the question is really asking! Study this graph.

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

ParabolaNormal

Tangent

Point (– 1, y1 )

The normal and tangent are at right angles to each other, so the gradient of the normal is found by first finding the gradient of the tangent then taking the NEGATIVE RECIPROCAL.

So, if the tangent’s gradient is 2/3, then the normal’s gradient will be – 3/2.

If tangent’s gradient is – 2, then normal’s will be ½ …etc

Questions requiring equations of TANGENTS AND NORMALS will almost always involve the formula

y – y1 = m(x – x1)You will need to substitute 3 constants into this equation……

x1, which will generally be given

y1, which will sometimes be given. If not, you can find it by subst. x1 into the original equation y = f(x) m, which usually is not given. You can find it by subst. x1 into the derived equation y = f ‘ (x), (i.e. subst into dy/dx)

Step 1 is to find y1 at the required point (-1, y1 ).

Subst. x = – 1 into y = x2 – 2x to get y = (– 1)2 – 2(– 1) = 3

Step 2 is to find m (i.e. y’) at the required point (-1, y1 ).

y1 = 3

y = x2 – 2x, so y’ = 2x – 2. Now subst x = – 1 into 2x – 2 to get y’ = – 4

m = – 4

y – y1 = m(x – x1)

1. TANGENT

y – 3 = – 4 (x + 1)i.e. y = – 4x – 1 Eq of tangent

Remember we need 3 constants: x1, y1 and m. We already know x1 = – 1, so we only have to find y1 and m.

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

ParabolaNormal

Tangent y = – 4x – 1

Point (– 1, 3)

Some observations so far…….

The ultimate formula you ALWAYS use in questions involving tangents and normals is y – y1 = m(x – x1). To use this effectively, you always need to find m, which you do by finding dx

dy

On some occasions (like this example) you may need to find y1 as well as m. This is done by substituting x1 (which you’re given) into the equation for y.

Step 1 is as before: find y1 at the required point (-1, y1 ).

Subst. x = – 1 into y = x2 – 2x to get y1 = (– 1)2 – 2(– 1) = 3

Step 2 is to first find mTAN (i.e. y’) at the required point (-1, y1 ) as we did before, then take negative reciprocal.

y1 = 3mtan = – 4

y – y1 = m (x – x1)

2. NORMAL

y – 3 = ¼ (x + 1)i.e. y = ¼ x + 3 ¼ Eq of normal

We use the 3 constants we found before, namely x1, y1 and m. But we have to take the NEGATIVE RECIPROCAL of m and use this as

our new m.

So mnorm = ¼

End of SectionClick to return to Menu

Section 2

Section 2. The Chain Rule – For differentiating expressions of the form y = (f [x])n

This form of differentiation enables you to deal with expressions that can’t be turned into polynomials and for which our formulae in Section 1 can’t be used. The rule you need to learn (at this present moment) is:

On Powerpoint #2 you will learn how to differentiate other functions including trigonometric and logarithmic functions. In that work, the chain rule is the dominant rule and it will assume different formats to 2.1 above.

2.1 )(')]([)]([ 1 xfxfnxfdx

d nn

Example 2.1 Differentiate y = (2x – 7)6

Solution If we chose to use the method of Section 1 we could expand this and then differentiate. But what a pain! So….enter the CHAIN RULE!!

2)72(6 5 xdx

dy

Here, f(x) = (2x – 7) and n = 6. Differentiating 2x – 6 gives f ’(x) = 2. So using the formula on the previous slide,

5)72(12 xdx

dyi.e. Ans

Example 2.2

Solution

4)54(2

1 21

x

dx

dy

Here, f(x) = (4x – 5) and n = ½ . Differentiating 4x – 5 gives f ’(x) = 4. So using the formula from 2 slides ago,

21

)54(2 x

dx

dyi.e.

Ans

Differentiate y = 54 x

First, rewrite as y = (4x – 5)1/2

or better still 54

2

xdx

dy

Example 2.3

Solution

xxdx

dy2)1(2 22

Here, f(x) = (x2 + 1) and n = – 1. Differentiating x2 + 1 gives f ’(x) = 2x. So using the formula from 3 slides ago,

22 )1(4 xxdx

dy

i.e.

Ans

Differentiate y = 1

22 x

First, rewrite as y = 2(x2 + 1)– 1

or better still22 )1(

4

x

x

dx

dy

NOTE the 2 in front just multiplies the n when you apply the formula. This gives the – 2 .

Example 2.4

Find the equation of the tangent to the curve 3

22

x

y

at the point where x = 1

Solution

Remember we need to use the formula y – y1 = m(x – x1)

and have to substitute 3 numbers: x1, y1 and m.

y1 = 1x1 = 1

3

22

x

yy1 is found by subst. x = 1 into

x1 is already given : it is 1.

232 )3(

2

x

x

dx

dym is found by subst. x = 1 into

m = -¼

First find2

32 )3(

2

x

x

dx

dydx

dy . Check that

Now we substitute these 3 constants into y – y1 = m(x – x1)

y – 1 = – ¼ (x – 1)

y – 1 = – ¼ x + ¼

y = – ¼ x + 1¼ is the equation of the tangent

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

Note (for interest only!) that the tangent line has a gradient of – ¼ and a y-intercept of 1 ¼ !


Section 3

Section 3. The Product Rule - for differentiating expressions of the form y = f(x) × g(x)

This method enables you to deal with expressions that are a multiplication of two other variable functions, i.e. of the form f(x) × g(x).

3.1 )()(')(')()]()([ xgxfxgxfxgxfdx

d

For convenience, the f(x) and g(x) are often written simply as f and g.

The product rule only need be used when the f(x) and g(x) are two distinct functions being multiplied together. Check first whether expanding might be a quicker and easier option! Also look out for invisible × signs !! BEWARE!!

Which of these would you need to use the product rule to differentiate?

)1()1()( 2 xxyi

52)( 3 xxyii

52 )1(4)( xxyiii

52 )1(4)( xyiv

64 )1()3()( xxyv

Easier to expand first then differentiate using basic polynomial rules. Product Rule could also be used with f = x – 1 and g = x2 + 1.

MUST use product rule with f = 2x and g = (x3 – 5)1/2

Cannot easily expand because of the power 5. Use Product Rule with f = 4x and g = (x2 + 1)5. To find g’ you will need to use the Chain Rule.

No need for Product Rule here as 4 is a constant. Use Chain Rule. Ans is 40x(x2 + 1)4

MUST use product rule with f = (x + 3)4 and g = (x – 1)6 . You will need to use Chain Rule to find both f’ and g’.

Note – There are invisible × signs in (ii), (iii) and (v)!

Example 3.1

Differentiate y = (x – 1)(x2 + 1) using the Product Rule

We differentiate each of these and set it out as follows. Easiest to set aside a workspace like this and get your four expressions ready to substitute:

f = x – 1

f ‘ = 1

g = x2 + 1 g‘ = 2x

Noting that f = x – 1 and g = x2 + 1

gffgdx

dy''

)1(12)1( 2 xxx

122 22 xxx

123 2 xx

Check that f ’ and g ’ are correct!

Note it would be quicker in this case to expand then differentiate! Expanding the expression gives x3 – x2 + x – 1, and its derivative is equal to 3x2 – 2x + 1 !!

Example 3.2

Differentiate y = (3x – 1)5 (x2 – 2)3 using the Product Rule

Noting that f = (2x – 1)5 and g = (x2 – 2)3 we differentiate these as follows using the Chain Rule in each case!

f = (3x – 1)5

f ‘ = 15(3x – 1)4

g = (x2 – 2)3 g‘ = 6x(x2 – 2)2

gffgdx

dy''

324225 )2()13(15)2(6)13( xxxxx324225 )2()13(15)2()13(6 xxxxx

You should now inspect these two terms to see if common factors can be taken out…..

324225 )2()13(15)2()13(6 xxxxx

To factorise

We “pair” the terms up and take out the highest common factor of each and position in front of the brackets below:

HCF of 6x and 15 is 3. Put 3 outside brackets and 2x & 5 inside

( )

HCF of (3x – 1)5 and (3x – 1)4 is (3x – 1)4. Put (3x – 1)4 outside brackets and (3x – 1) & 1 inside

3 2x +5(3x – 1)4 (3x – 1) (1)(x2 – 2)2 (1) (x2 – 2)

HCF of (x2 – 2)2 and (x2 – 2)3 is (x2 – 2)2. Put (x2 – 2)2 outside brackets and (1) and (x2 – 2) inside.

Now expand and clean up the expressions in the large brackets:

= 3(3x – 1)4(x2 – 2)2 (11x2 – 2x – 10) ANS

Example 3.3

We differentiate each of these as follows.

f = 2x

f ‘ = 2

g = (x2 – 1)1/2 g‘ = x(x2 – 1) – 1/2

Noting that f = 2x and g = (x2 – 1)1/2

gffgdx

dy''

2122

12 )1(2)1(2 xxxx

g ’ is obtained by Chain Rule – REMEMBER??

Differentiate using the Product Rule12 2 xxy

121

2 2

2

2

xx

xAns….but see over

121

2 2

2

2

xx

xIt is worth noting that expressions like

can be further simplified

121

2 2

2

2

xx

x

We use the result b

bca

b

bc

b

ac

b

a

1

112

1

22

22

2

2

x

xx

x

x

1

)1(222

22

x

xx

1

242

2

x

x

a

b

c

Check this!!

Noting that the square root signs multiply out. i.e. a a = a


Section 4

This method enables you to deal with expressions that are FRACTIONS with variable functions in top & bottom, i.e. of the form f(x) / g(x).

For convenience, again the f(x) and g(x) can be written as f and g.

The Quotient Rule need only be used when the fraction cannot be simplified and put into axn format (like Example 1.4). Usually fractions with denominators containing more than one term would indicate the need for the Quotient Rule to be used.

Section 4. The Quotient Rule – for differentiating expressions of the form

)(

)(

xg

xfy

2)]([

)()(')(')(

)(

)(

xg

xfxgxfxg

xg

xf

dx

d

NOTE! The order of terms in the top is important. The format is like the product rule but because of the minus you must begin with gf’

Which of these would you need to use the Quotient Rule to differentiate?

x

xy

12

No need for QR as it can be broken into and differentiated easily using axn rules.

1 xxy

2

12

x

xy QR necessary here. See example 4.1 next slide

3

92

x

xy

Hmmm…Depends whether you’re awake or not! The x2 – 9 factorises into (x – 3)(x + 3) so the whole thing just is equal to x – 3 which differentiates to 1 ! Otherwise use QR.

1

x

xy QR necessary here. See example 4.2 next slides

Example 4.1

Differentiate using the Quotient Rule2

12

x

xy

f = x2 – 1

f ‘ = 2x

g = x + 2g‘ = 1

Again like the Product Rule, set out your component parts as follows:

2

''

g

fggf

dx

dy

2

2

)2(

)1)(1()2)(2(

x

xxx

2

2

)2(

14

x

xx

Note – the numerator is usually expanded & simplified, but the denominator is left in factorised format.

Example 4.2

Differentiate using the Quotient Rule1

x

xy

f = x

f ‘ = 1

g = x + 1g‘ = 1

Setting out parts as follows:

2

''

g

fggf

dx

dy

2)1(

))(1()1)(1(

x

xx

2)1(

1

x

Note – again the numerator is expanded & simplified, but the denominator is left in factorised format.

Example 4.3

Differentiate 1+

=x

xy

Setting out parts as follows:

xf =

xf

2

1='

1+= xg

1+2

1='

xg Chain rule

here

See slides 15, 17, 18 to revisit differentiation of surds!

2

''

g

fggf

dx

dy

1

12

1

2

11

x

xxx

x

1122

1

x

x

x

x

x

Now work on simplifying the numerator (see over)

122

1

x

x

x

x To simplify this expression, we use basic fraction subtraction rules i.e.

14

2)1(2

xx

xx

xx

22

1

bd

bcad

d

c

b

a

This becomes the

top of the original

expression

1122

1

x

x

x

x

x

12

12

xxx

xxx

2)1(2

1

Ans

Note! This last step uses the rule

bc

a

cb

a

The whole expression becomes simpler if we can first simplify the numerator!

Now the original….

You might find this helpful for remembering the Quotient Rule

When you begin writing the Quotient Rule, begin with

2

...............

g

g

dx

dy

Now just complete the rest of the top as if it were the Product Rule, but REMEMBER TO PUT A MINUS BETWEEN THE TERMS, NOT A PLUS!!

f’ – g’ f

END SHOW

How can I subtract 1 from any fraction quickly and easily?

14

3

Here’s the rule….Subtract the bottom from the top and this becomes the new top! So, to do

We calculate 3 – 4, get the result – 1 and this becomes the top of our answer 4

1

i.e.4

1

17

5

7

2Think

5 – 7 = – 2 7

2

NOTE! A minus in the top (or bottom) of a fraction is the same as if it were right out the front

14

31

4

3

– 3 – 4 = – 7 4

7

4

7

RETURN

introduction to differentiation i basic functions f servello

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