Introduction to
Differentiation I
Basic 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…..
y
f(x)
Then the Derived
function should
be called…..
dy
y’ or
dx
f ’ or f ’ (x)
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
2
x2  3
3 sin x
4x
5x2 – 3 tan2x
3x  4
*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
1.2
1.3
x is the variable;
d
n
n 1
(ax )  nax
whereas a and n
are constants
dx
d
This is just a special
(ax)  a
case of 1.1 with n = 1.
dx
So 3x becomes 3; – 2x
becomes – 2 etc…
d
(any const )  0
dx
Another special case of 1.1 with
n = 0. Constants (numbers on their
own) always just become 0.
Section 1: Examples to try
Click to check your answers!
1
2
3
4
d
Note the 3 and 5 are multiplied, and the
5
4
(3 x )  15x
original power (5) is lowered by 1 to give 4
dx
d 2
2 is treated as 1x2 so it
Note
x
( x  5 x)  2x – 5
becomes 2x1 i.e. 2x, and 5x just
dx
becomes 5.
d
Note the 3x becomes 3, and the – 7
( 2 x 2  3 x  7) 
disappears because it’s a constant.
dx
4x – 3
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 dy
dx
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
CAUTION! If the power
were higher than a ”square”,
expanding might be too
onerous! There is an
alternative we learn later
(the Chain Rule)
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
(a – b)2 = a2 – 2ab + b2
format and so it’s differentiable!
Using our rules
Answer: f ’(x) = 18x – 30 from before!
Expressions which can be put into polynomial
format, i.e. axn.
Example 1.2
dy
If y = (x – 7)(x + 7), find
dx
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
Which is now in polynomial
format and hence differentiable!
Answer: y’ = 2x
Did you remember the
difference of 2 squares?
(a – b)(a + b) = a2 – b2
Using our rules
from before!
Background info slide #3
When multiplying a whole number or letter by a fraction, the
whole number multiplies into the TOP of the fraction!
2 6
3 
5 5
5x 2
5
 x2
9
9
3 3x
x 
7
7
Ability to manipulate these expressions will be
very handy as we move into more difficult
examples in later slides.
3
5
Note 8  
Example
Solution
Differentiate
3x8
5
3 8
3x8
Use (3) above to rewrite
as x
5
5
dy
24 7
Differentiating gives
= x
dx
5
24
5
Ans!
Expressions which can be put into polynomial
format, i.e. axn.
Example 1.3
dy
3x
Given y =
, find
dx
5
Solution
3x
We need to get the into the format axn.. This
5
requires a simple manipulation from
3x
3
to x which is now of the 3
5
5
format axn [Note a =
5
& n = 1]
Answer:
dy 3

dx 5
Expressions which can be put into polynomial
format, i.e. axn.
Example 1.4
Given y =
Solution
2 x 4  3x 3find dy
dx
2
5x
Decompose the fraction as follows:
2 x 4  3x 3 2 x 4 3x 3
Now we can differentiate!


5x 2 5x 2
5x 2
2 2 3
y

x  x
2
2 x 3x
5
5


5
5
dy 4
3
2 2 3
 x
Ans:
 x  x
dx 5
5
5
5
which is now in
diffrentiable format!
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
1
n

x
xn
xx
1
n
2
xx
1
n
Make sure you can do manipulations like these…..
2
3

2
x
x3
5
5 7
 x
7
3x
3
3
2
2
Study this one closely!
2x x  2x  x
1 1 1
 x
2x 2
5 x  5x
1
2
2
1
 2x
2  12
 x
7 x 7
2 x
3
Example 1.5
Solution
1
Differentiate
x6
1
6
Rewrite y  6 as y  x
x
dy
6
dy
 7
So
or
 6 x  7
dx
x
dx
Ans
Example 1.6
Solution
Differentiate
2
7 x3
2
2 3
y x
Rewrite y 
3 as
7x
7
dy
6
dy
6

So
or
4
  x 4
dx
7
x
dx
7
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
Solution
Differentiate y  6
Rewrite y  6
So dy  3 x  1 2
dx
x
x as y  6x
i.e.
1
2
dy
3

dx
x
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
Solution
Differentiate
Rewrite as
f ( x) 
3
2 x
 43 x
1
3  12
y x
 4x 3
2
Which is now in differentiable axn format
dy
3  32 4 23
 x  x
dx
4
3
Ans
1 3
3
  
2 2
4
OR…
dy
3
4


dx
4 x 3 3.3 x 2
Make sure you can
follow this!
NOTE!! It may be timely here to
revise how to multiply fractions. Just
multiply across the top and then across
the bottom. So
Ans
DID YOU KNOW? There’s
a fast way to subtract 1
from any fraction (in your
head) !
SHOW ME!
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)
x
2
(b)
3
(c)
1
5
x  x x  x
1
7
7
2
7
2
(x
)

x
x 
2
2
2
2
1
2
Add powers when
multiplying: 2 + ½ = 5/2
Multiply powers when
removing brackets
2
2  53
2 x
2x 3
2( x )
x



7
1
2 3
2
5
5 x . x 5x  x 3
5x 3
3
Subtract powers when
dividing
Example
1.9
Solution
3x 2  2 x x
y
5 x.3 x
Differentiate
3x 2
2x x
Split first: y 

3
5 x. x
5 x.3 x
Change surds to
powers:
3x
y
2
5 x.x
Simplify each top
& bottom by
adding powers:
y
1
3x

3
2
4
3
2 x.x
5 x.x

2x
3
1
1
2
3
2
4
5x
5x 3
3 23
2 16
y x  x
5
5
Extract fractions to
front & subtract
powers
Which is now differentiable!
3 23
2 16
y x  x
5
5
dy
2  13
1 56
 x

x
dx
5
15
Ans
Remember – it’s a good idea to practise turning this index
notation back into surd format
dy
2
1
 3

dx
5. x 15.6 x 5
Ans
Example
1.10
Find the equation of the tangent and
normal to the parabola y = x2 – 2x at
the point where x = – 1.
y
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
5.0
Point (– 1, y1 )
4
3
Normal
2
Parabola
1
x
-5.0
-4
-3
-2
-1
1
2
3
4
-1
-2
-3
-4
-5.0
Tangent
5.0
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)
1. TANGENT
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
Some observations so far…….
The ultimate formula you
ALWAYS use in questions
Point (– 1, 3)
involving tangents and
normals is y – y1 = m(x
– x1). To use this
Normal
effectively, you always
need to find m, which you
dy -5.0 -4 -3 -2
do by finding
dx
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.
y
5.0
4
3
2
Parabola
1
x
-1
1
2
3
4
5.0
-1
-2
-3
-4
-5.0
Tangent
y = – 4x – 1
2. 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.
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
y1 = 3
End of Section
Click to return to Menu
Step 2 is to first find mTAN
(i.e. y’) at the required point
(-1, y1 ) as we did before, then
take negative reciprocal.
mtan = – 4
So mnorm = ¼
y – y1 = m (x – x1)
y – 3 = ¼ (x + 1)
i.e. y = ¼ x + 3 ¼
Eq of normal
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:
2.1
d
[ f ( x)]n  n[ f ( x)]n 1  f ' ( x)
dx
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.
Example 2.1
Solution
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!!
Here, f(x) = (2x – 7) and n = 6.
Differentiating 2x – 6 gives f ’(x) = 2. So
using the formula on the previous slide,
dy
 6(2 x  7) 5  2 i.e.
dx
dy
5
 12(2 x  7) Ans
dx
Example 2.2
Differentiate y =
Solution
4x  5
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,
dy
1
i.e.
 2(4 x  5) 2
dy 1
1
dx
2
 (4 x  5)  4
dy
2
dx 2

dx
or better still
4x  5
Ans
Example 2.3
Solution
Differentiate y =
2
x2 1
First, rewrite as y = 2(x2 + 1)– 1
Here, f(x) = (x2 + 1) and n = – 1.
Differentiating x2 + 1 gives f ’(x) = 2x. So
using the formula from 3 slides ago,
dy
 2( x 2  1)  2  2 x
dx
dy
 4 x( x 2  1)  2
dx
or better still
i.e.
NOTE the 2 in front just
multiplies the n when you
apply the formula. This gives
the – 2 .
dy
4x
 2
2
dx
( x  1)
Ans
Example 2.4
Find the equation of the tangent to the curve
at the point where x = 1
Solution
dy
First find .
dx
Check that
y
x1 is already given : it is 1.
y1 is found by subst. x = 1 into
m is found by subst. x = 1
x2  3
dy
2x

3
2
dx
( x  3) 2
Remember we need to use the formula y – y1
and have to substitute 3 numbers: x1, y1 and m.
y
2
2
x2  3
2x
dy
3
into dx   2
( x  3) 2
= m(x – x1)
x1 = 1
y1 = 1
m = -¼
Now we substitute these 3 constants
into y – y1 = m(x – x1)
y – 1 = – ¼ (x – 1)
y
y–1= –¼x+¼
5.0
4
y = – ¼ x + 1¼
3
2
is the equation of the tangent
1
x
-5.0
-4
-3
-2
-1
1
-1
Note (for interest only!) that the
tangent line has a gradient of – ¼
and a y-intercept of 1 ¼ !
-2
-3
-4
End of Section
Click to return to Menu
-5.0
2
3
4
5.0
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
d
[ f ( x)  g ( x)]  f ( x) g ' ( x)  f ' ( x) g ( x)
dx
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?
(i ) y  ( x  1)  ( x  1)
2
(ii ) y  2 x x 3  5
(iii ) y  4 x( x 2  1) 5
(iv ) y  4( x 2  1) 5
(v) y  ( x  3) 4 ( x  1) 6
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
Noting that f = x – 1 and g = x2 + 1
f=x–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:
f‘=1
dy
 fg ' f ' g
dx
 ( x  1)  2 x  1 ( x 2  1)

 2x2  2x  x2 1
 3x 2  2 x  1
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 !!
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!
dy
 fg ' f ' g
dx
f = (3x – 1)5
f ‘ = 15(3x – 1)4
g = (x2 – 2)3
g‘ = 6x(x2 – 2)2
 (3x  1)5  6 x( x 2  2) 2  15(3x  1) 4  ( x 2  2)3
 6 x(3x  1)5 ( x 2  2) 2  15(3x  1) 4 ( x 2  2)3
You should now inspect these two terms to
see if common factors can be taken out…..
To factorise
6 x(3x  1)5 ( x 2  2) 2  15(3x  1) 4 ( x 2  2)3
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.
2
2
+5 (1) (x2 – 2))
3 (3x – 1)4 (x – 2) ( 2x(3x – 1) (1)
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
Differentiate y  2 x x 2  1 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
dy
 fg ' f ' g
dx
 2 x  x( x  1)
2

2x2
x2 1
g ’ is obtained by Chain
Rule – REMEMBER??
1
2
 2  ( x  1)
2
1
2
 2 x 2  1 Ans….but see over
It is worth noting
that expressions
like
We use the result
2x2
x2 1
 2 x2 1
can be further
simplified
a
a bc a  bc
c  

b
b b
b
a
2x2
x 1
2
 2 x 1 
b
2x2
2
c

x 1
x2 1
2 x 2  2( x 2  1)

End of Section
Click to return to Menu

2 x2 1 x2 1
2
x 1
2
4x2  2
x2 1
Check
this!!
Noting that the square
root signs multiply out.
i.e. a a = a
Section 4
Section 4. The Quotient Rule – for differentiating
f ( x)
expressions of the form y 
g ( x)
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
d  f ( x)  g ( x) f ' ( x)  g ' ( x) f ( x)

 
dx  g ( x) 
[ g ( x)]2
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.
Which of these would you need to use
the Quotient Rule to differentiate?
x 1
y
x
No need for QR as it can be broken into y  x  x
and differentiated easily using axn rules.
x2 1
y
x2
QR necessary here. See example 4.1 next slide
2
x 9
y
x3
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.
x
y
x 1
QR necessary here. See example 4.2 next slides
2
1
Example 4.1
x2 1
Differentiate y 
using the Quotient Rule
x2
2
f=x –1
Again like the Product Rule,
set out your component parts
as follows:
dy gf ' g ' f

dx
g2
( x  2)( 2 x)  (1)( x 2  1)

( x  2) 2
x2  4x 1

( x  2) 2
f ‘ = 2x
g=x+2
g‘ = 1
Note – the numerator is usually
expanded & simplified, but the
denominator is left in factorised
format.
Example 4.2
x
Differentiate y 
using the Quotient Rule
x 1
f=x
Setting out parts as follows:
dy gf ' g ' f

dx
g2
( x  1)(1)  (1)( x)

( x  1) 2
1

( x  1) 2
f‘=1
g=x+1
g‘ = 1
Note – again the numerator is
expanded & simplified, but the
denominator is left in factorised
format.
Example 4.3
Differentiate y =
x
x +1
f = x
1
f '=
2 x
Setting out parts as follows:
dy gf ' g ' f

dx
g2



 
 1   1 
x 1 

 x
 2 x   2 x 1 
x 1
x 1
x

2 x 1
 2 x
x 1
g = x +1
1
g'=
2 x +1
Chain rule
here
See slides 15, 17, 18 to
revisit differentiation of
surds!
Now work on simplifying
the numerator (see over)
The whole expression becomes simpler if we can first simplify
the numerator!
x 1
x

2 x 2 x 1
2( x  1)  2 x

4 x x 1

To simplify this expression, we use basic
fraction subtraction rules i.e.
a c ad  bc
 
b d
bd
1
Note!
This last
step uses
the rule
2 x x
2
x 1
x

Now the
2 x 1
 2 x
original….
x 1
1
 2 x x
x 1
2

a
b a
c
bc
1
2( x  1) x 2  x
Ans
You might find this helpful for remembering
the Quotient Rule
When you begin writing the Quotient Rule,
begin with
dy g..........
f’ – g’ f.....

dx
g2
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
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
3
1
4
1
We calculate 3 – 4, get the result – 1
and this becomes the top of our answer 4
1
i.e. 
4
5
1 
7
Think
5–7=–
2
NOTE! A minus in the top (or
bottom) of a fraction is the same
as if it were right out the front
2
2

7
7
RETURN
3
3
7
7
 1 
1 

4
4
4
4
–3 –4= –7