1. Instantaneous rate of change
Introduction to
Differentiation I
Basic Functions
Instantaneous rate of change
dy/dx
f'(x)
Gradient of tangent
F Servello

2. 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!

3. Differentiating Polynomials
Section 1
Differentiating Polynomials

4. 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’ or
f(x)
f ’
f ’ (x)

5. Examples of “polynomials” (for our purposes*) include:
Background info slide #2
For 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 8
Examples of non-polynomials (for our purposes*) include:
(4x – 7)8 3 sin x 4x
5x2 – 3 tan2x
*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.

6. Section 1. Differentiating Polynomials: axn
This is the most basic form of differentiation. The rules you must learn are as follows:
1.1
x is the variable; whereas a and n are constants
1.2
This is just a special case of 1.1 with n = 1. So 3x becomes 3; – 2x becomes – 2 etc…
1.3
Another special case of 1.1 with n = 0. Constants (numbers on their own) always just become 0.

7. Section 1: Examples to try
Click to check your answers! 1
Note the 3 and 5 are multiplied, and the original power (5) is lowered by 1 to give 4
15x4 2
2x – 5
Note x2 is treated as 1x2 so it becomes 2x1 i.e. 2x, and 5x just becomes 5. 3
Note the 3x becomes 3, and the – 7 disappears because it’s a constant.
4x – 3 4
If y = 3x2 – ½ x4 – x , find y’.
Note – the alternative phrasing of this question.
y’ merely means
Ans: y’ = 6x – 2x3 – 1 , using previous rules 1.1, 1.2

8. "Disguised" Polynomials

9.  x  ±  x  ±  x  ….etc Important note
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

10. “Disguised polynomials” : Expressions which can be put into polynomial format, i.e. axn.
Example 1.1
CAUTION! If the power were higher than a ”square”, expanding might be too onerous! There is an alternative we learn later (the Chain Rule)
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?
Which is now in polynomial format and so it’s differentiable!
(a – b)2 = a2 – 2ab + b2
Using our rules from before!
Answer: f ’(x) = 18x – 30

11. Expressions which can be put into polynomial format, i.e. axn.
Example 1.2
If y = (x – 7)(x + 7), find
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?
Which is now in polynomial format and hence differentiable!
(a – b)(a + b) = a2 – b2
Using our rules from before!
Answer: y’ = 2x

12. Ans! Background info slide #3
When multiplying a whole number or letter by a fraction, the whole number multiplies into the TOP of the fraction!
Ability to manipulate these expressions will be very handy as we move into more difficult examples in later slides.
Note
Differentiate
Example
Solution
Use (3) above to rewrite as
Differentiating gives =
Ans!

13. Answer: Expressions which can be put into polynomial format, i.e. axn.
Example 1.3
Given y = , find
Solution
We need to get the into the format axn.. This requires a simple manipulation from
which is now of the format axn [Note a = & n = 1] to
Answer:

14. Ans: Expressions which can be put into polynomial format, i.e. axn.
Example 1.4
Given y = find
Solution
Decompose the fraction as follows:
Now we can differentiate!
Ans:
which is now in diffrentiable format!

15. Make sure you can do manipulations like these…..
Background info slide #4
Negative and Fractional Powers can be very useful in differentiating! Learn these rules and refer to the INDICES Powerpoint for further examples
Make sure you can do manipulations like these…..
Study this one closely!

16. Ans Ans Differentiate Example 1.5 as Rewrite Solution So or

17. Important note 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.

18. Ans Example 1.7 Differentiate as Rewrite Solution So i.e.
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)

19. Which is now in differentiable axn format
Example 1.8
Differentiate
Rewrite as
Solution
Which is now in differentiable axn format
NOTE!! It may be timely here to revise how to multiply fractions. Just multiply across the top and then across the bottom. So
Ans
OR…
DID YOU KNOW? There’s a fast way to subtract 1 from any fraction (in your head) !
SHOW ME!
Make sure you can follow this!
Ans

20. Background info slide #5
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
(a)
Add powers when multiplying: 2 + ½ = 5/2
(b)
Multiply powers when removing brackets
(c)
Subtract powers when dividing

21. Example 1.9
Differentiate
Solution
Split first:
Change surds to powers:
Simplify each top & bottom by adding powers:
Extract fractions to front & subtract powers
Which is now differentiable!

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

23. Find the equation of the tangent and normal to the parabola y = x2 – 2x at the point where x = – 1.
Example 1.10
Parabola
Normal
Tangent
Point (– 1, y1 )
It may be useful to know what the question is really asking! Study this graph.
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

24. y – y1 = m(x – x1) Important note
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)

25. 1. TANGENT y – y1 = m(x – x1) y – 3 = – 4 (x + 1) i.e. y = – 4x – 1
Remember we need 3 constants: x1, y1 and m. We already know x1 = – 1, so we only have to find y1 and m.
Step 1 is to find y1 at the required point (-1, y1 ).
Step 2 is to find m (i.e. y’) at the required point (-1, y1 ).
Subst. x = – 1 into y = x2 – 2x to get y = (– 1)2 – 2(– 1) = 3
y = x2 – 2x, so y’ = 2x – 2. Now subst x = – 1 into 2x – 2 to get y’ = – 4
y1 = 3
m = – 4
y – y1 = m(x – x1)
y – 3 = – 4 (x + 1)
i.e. y = – 4x – 1
Eq of tangent

26. 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
Parabola
Normal
Tangent y = – 4x – 1
Point (– 1, 3)
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.

27. 2. NORMAL y – y1 = m (x – x1) y – 3 = ¼ (x + 1) i.e. y = ¼ x + 3 ¼
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.
Step 1 is as before: find y1 at the required point (-1, y1 ).
Step 2 is to first find mTAN (i.e. y’) at the required point (-1, y1 ) as we did before, then take negative reciprocal.
Subst. x = – 1 into y = x2 – 2x to get y1 = (– 1)2 – 2(– 1) = 3
mtan = – 4
y1 = 3
So mnorm = ¼
y – y1 = m (x – x1)
y – 3 = ¼ (x + 1)
End of Section
Click to return to Menu
i.e. y = ¼ x + 3 ¼
Eq of normal

28. Section 2
The Chain Rule

29. 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:
2.1
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.

30. Example 2.1
Differentiate y = (2x – 7)6
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!!
Solution
Here, f(x) = (2x – 7) and n = 6. Differentiating 2x – 6 gives f ’(x) = 2. So using the formula on the previous slide,
i.e.
Ans

31. Example 2.2
Differentiate y =
Solution
First, rewrite as y = (4x – 5)1/2
Here, f(x) = (4x – 5) and n = ½ . Differentiating 4x – 5 gives f ’(x) = 4. So using the formula from 2 slides ago,
i.e.
Ans
or better still

32. Differentiate y =
Example 2.3
Solution
First, rewrite as y = 2(x2 + 1)– 1
NOTE the 2 in front just multiplies the n when you apply the formula. This gives the – 2 .
Here, f(x) = (x2 + 1) and n = – 1. Differentiating x2 + 1 gives f ’(x) = 2x. So using the formula from 3 slides ago,
i.e.
Ans
or better still

33. Example 2.4
Find the equation of the tangent to the curve
at the point where x = 1
Solution
First find
Check that
Remember we need to use the formula y – y1 = m(x – x1) and have to substitute 3 numbers: x1, y1 and m.
x1 = 1
x1 is already given : it is 1.
y1 = 1
y1 is found by subst. x = 1 into
m = -¼
m is found by subst. x = 1 into

34. 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
Note (for interest only!) that the tangent line has a gradient of – ¼ and a y-intercept of 1 ¼ !
End of Section
Click to return to Menu

35. Section 3
The Product Rule

36. 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
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!!

37. Which of these would you need to use the product rule to differentiate?
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)!

38. Differentiate y = (x – 1)(x2 + 1) using the Product Rule
Example 3.1
Differentiate y = (x – 1)(x2 + 1) using the Product Rule
Noting that f = x – 1 and g = x2 + 1
f = x – 1
f ‘ = 1
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:
g = x2 + 1
g‘ = 2x
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 !!

39. 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
You should now inspect these two terms to see if common factors can be taken out…..

40. ( ) = 3(3x – 1)4(x2 – 2)2 (11x2 – 2x – 10) ANS 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
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. ( ) 3 2x +5
(3x – 1)4
(3x – 1)
(1)
(x2 – 2)2
(1)
(x2 – 2)
Now expand and clean up the expressions in the large brackets:
= 3(3x – 1)4(x2 – 2)2 (11x2 – 2x – 10) ANS

41. Differentiate using the Product Rule
Example 3.3
Differentiate using the Product Rule
Noting that f = 2x and g = (x2 – 1)1/2
f = 2x
f ‘ = 2
We differentiate each of these as follows.
g = (x2 – 1)1/2
g‘ = x(x2 – 1) – 1/2
g ’ is obtained by Chain Rule – REMEMBER??
Ans….but see over

42. It is worth noting that expressions like can be further simplified
We use the result a b c
Noting that the square root signs multiply out. i.e. a a = a
Check this!!
End of Section
Click to return to Menu

43. Section 4
The Quotient Rule

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

45. Which of these would you need to use the Quotient Rule to differentiate?
No need for QR as it can be broken into and differentiated easily using axn rules.
QR necessary here. See example 4.1 next slide
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.
QR necessary here. See example 4.2 next slides

46. Differentiate using the Quotient Rule
Example 4.1
Differentiate using the Quotient Rule
f = x2 – 1
Again like the Product Rule, set out your component parts as follows:
f ‘ = 2x
g = x + 2
g‘ = 1
Note – the numerator is usually expanded & simplified, but the denominator is left in factorised format.

47. Differentiate using the Quotient Rule
Example 4.2
Differentiate using the Quotient Rule
f = x
f ‘ = 1
Setting out parts as follows:
g = x + 1
g‘ = 1
Note – again the numerator is expanded & simplified, but the denominator is left in factorised format.

48. Setting out parts as follows:
Example 4.3
Differentiate
Setting out parts as follows:
Chain rule here
See slides 15, 17, 18 to revisit differentiation of surds!
Now work on simplifying the numerator (see over)

49. The whole expression becomes simpler if we can first simplify the numerator!
To simplify this expression, we use basic fraction subtraction rules i.e.
Note! This last step uses the rule
This becomes the top of the original expression
Now the original….
Ans

50. You might find this helpful for remembering the Quotient Rule
When you begin writing the Quotient Rule, begin with
f’ – g’ f
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!!
END SHOW

51. How can I subtract 1 from any fraction quickly and easily?
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
i.e.
NOTE! A minus in the top (or bottom) of a fraction is the same as if it were right out the front
Think – 7 = – 2
RETURN
– 3 – 4 = – 7