LO1 – Differentiation Pupil sheets – note and lesson

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Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
CALCULUS
–
DIFFERENTIATION
What do these pieces of notation mean:








y
dy
and
and f ( x)
x
dx
and
 2y
d2y
and
and f ( x)
 x2
dx 2
What is derivative?
What does it mean to differentiate?
How do we differentiate?
What is the derivative of:
(a) y  sin x
derivative
rate of change
formula for gradient of a tangent to a curve
velocity (where x is distance and y is time)
the rate at which the rate of change is changing
second derivative
acceleration (where x is distance and y is time)
formula for determining the nature of a stationary point
and
(b)
y  cos x ?
Differentiation from First Principles
Let us differentiate y  x 2 from first principles.
The derivative represents the instantaneous gradient at any given point.
y
To find the gradient at the point (x, x2) we can
take a point nearby, (x + h, (x + h)2) and
calculate the gradient between these points.
The closer the points are the better our result
i.e. h  0 our calculation improves.
(x + h)2
x2
x
f ( x)  lim
x 0
 lim
x 0
 lim
x 0
 lim
x 0
 2x
Page 1
x
x+h
( x  h) 2  x 2
( x  h)  x
x 2  2 xh  h 2  x 2
h
2
2 xh  h
h
2x  h
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
Chain Rule (differentiating a function of a function)
If
f(x) = g(h(x))
then
or, in Leibniz notation, if
f’(x) = g’(h(x)).h’(x)
y = g(h(x))

then let u = h(x)
dy dy du
.


dx du dx
The unofficial Chain Rule is “Differentiate the outer function then multiply by the derivative of the
inner function”.
This can easily be expanded to more complex functions:
if
y = g(h(k(x)))
then let u = k(x)
 y = g(h(u))
and let v = h(u)
 y = g(v)
dy dy dv du



dx dv du dx
hence,
Examples:
1.
Differentiate y = (3x2 + 4x + 5)3.

Let u = 3x2 + 4x + 5  y = u3
dy dy du


dx du dx
 3u 2  (6 x  4)
 3(3x 2  4 x  5)(6 x  4)
 6(3x 2  4 x  5)(3x  2)
2.
Calculate
dy
when y = sin2(3x).
dx
Let u = 3x
 y = sin2u
dy dy dv du



dx dv du dx
 2v  cos u  3
 2(sin u)  cos u  3
 6  sin 3x  cos 3x
 3(2sin 3x  cos 3x)
 3sin 6 x
3.
Complete
  15v 2  2v7 


v 
3v 4

Page 33, Exercise 3B – questions 1(b, a), 2(b, c), 4, 3c
Page 2
and let v = sin u
 y = v2
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
There can be some confusion with the notations for inverse trigonometric functions.
a 1 
1
.
a
We know that
a2  a  a
We also know that
sin 2 x  (sin x)2  (sin x)  (sin x)
This implies that
sin 1 x  (sin x) 1 
and
1
sin x
so therefore the power for a
trigonometric function comes after the
ratio and before the algebraic term.
but we know that this is not the case.
[N.B. In many countries (and much of Britain is trying to implement this change) the inverse
trigonometric functions are called arcsine, arccosine and arctangent.]
To avoid this confusion, we denote functions as follows:
inverse sine as sin 1 x , inverse cosine as cos 1 x and inverse tangent as tan 1 x , just as we have
always done.
1
 cosec x
sin x
1
 sec x
cos x
1
 cot x
tan x
However, we will denote the following functions as:
Using the Chain Rule, find:
dy
4.
when y  cosec x where x is defined on a suitable domain.
dx
y  cosec x 
5.
Page 3
1
 (sin x) 1
sin x

dy

dx
dy
when y  sec x where x is defined on a suitable domain.
dx
i.e. cosecant
i.e. secant
i.e. cotangent.
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
The Product Rule
If
f(x) = g(x).h(x)
then
f’(x) = g’(x).h(x) + g(x).h’(x)
or, using Leibniz notation, if
y = uv
then
dy du

v  u dv
dx dx
dx
[N.B. we do “derivative of 1st times 2nd untouched plus 1st untouched times derivative of 2nd”.]
Proof
f ( x)  u ( x)  v( x)
f ( x  h)  f ( x)
h
u( x  h)  v( x  h)  u ( x)  v( x)
 lim
h 0
h
u( x  h)  v( x  h)  u ( x)  v( x  h)  u ( x)  v( x  h)  u ( x)  v( x)
 lim
h 0
h
v ( x  h)  u ( x  h )  u ( x )   u ( x )  v ( x  h)  v ( x ) 
 lim
h 0
h
v ( x  h)  u ( x  h )  u ( x ) 
u ( x )  v ( x  h)  v ( x ) 
 lim
 lim
h 0
h

0
h
h
u( x  h)  u( x)  lim u( x) v( x  h)  v( x)
 lim
v( x  h)
h 0
h0
h
h
 v( x)  u '( x)  u ( x)  v '( x)
f '( x)  lim
h 0
“central” terms introduced
for convenience
 u '( x)  v( x)  u( x)  v '( x)
Examples:
6.
Differentiate y = (2x – 3)(3x + 5).
dy
 (2)(3x  5)  (2 x  3)(3)
dx
 6 x  10  6 x  9
 12 x  1
7.
Calculate
dy
when y = sin2x . cos2x where x is defined on a suitable domain.
dx
dy
 d  sin 2 x   cos 2 x  sin 2 x  d  cos 2 x 
dx dx
dx
2
  2sin x  cos x   cos 2 x  sin x   2sin 2 x 
 2sin x cos x cos 2 x  2sin 2 x sin 2 x
 sin 2 x cos 2 x  2sin x sin 2 x
2
 sin 2 x  cos 2 x  2sin 2 x 
Page 4
[ N .B. 2sin x cos x  sin 2 x]
Bridge of Don Academy – Department of Mathematics
8.
Mr B. A. Willox
Calculate f’(x) when f(x) = (x2 + 2)(x – 1)2.
f '( x)  2 x  ( x  1) 2  ( x 2  2)  2( x  1)
 2( x  1)  x( x  1)  ( x 2  2)
 2( x  1)(2 x 2  x  2)
9.
Calculate
 3x 2  3 
dy
when y  sin 2 3x  
 where x is defined on a suitable domain.
dx
x


Page 35, Exercise 4A – questions 1(a, c, e), 3(b, h, j)
Page 36, Exercise 4B – questions 1(b, c), 5, 6
Page 5
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
The Quotient Rule
If f ( x) 
g ( x)
then
h( x)
f '( x) 
g '( x)  h( x)  g ( x)  h '( x)
(h( x)) 2
dy

or, using Leibniz notation, if y  u then
dx
v
 dudx  v  u  dvdx 
v2
[N.B. derivative of top  bottom – top  derivative of bottom, all over bottom squared.]
f ( x) 
Proof
1
g ( x)
 g ( x)   h( x) 
h( x)
f '( x)  g '( x)   h( x)   g ( x)   h( x)   h '( x)
1
2
using the Product Rule
and the Chain Rule
h '( x)
 g '( x)  1  g ( x) 
2
h( x)
 h( x) 
 g '( x) 

h( x)
h '( x)
2  g ( x) 
2
 h( x) 
 h( x) 
g '( x)  h( x)  g ( x)  h '( x)
2
 h( x) 
Examples:
10.
2
Differentiate y  x  3x  5 .
x2
dy (2 x  3)( x  2)  ( x 2  3x  5)(1)

dx
( x  2) 2
2
2
 2 x  7 x  6  x2  3x  5
( x  2)
2
 x  4 x 2 1
( x  2)
11.
2


Calculate d  5x
.
dx  3  2 x 

d  5x 2   d 
5x 2
dx  3  2 x  dx   3  2 x  1 2





1
1
10 x  (3  2 x) 2  5 x 2  1 (3  2 x) 2  (2)
2

2
1
3  2x  2

10 x  3  2 x  5 x 2 

Page 6

1
3  2x
3  2x
1
10 x  3  2 x  5 x 2 
3  2x
3  2x

3  2x
10 x(3  2 x)  5 x 2

3  2 x  (3  2 x)
15 x(2  x)

3
3  2x  2
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
12. IMPORTANT!! Calculate the derivative of y = tan x.
y  tan x  sin x
cos x
dy cos x  cos x  sin x  ( sin x)

dx
(cos x) 2
2
2
 cos x 2 sin x
cos x
 12
cos x
 sec 2 x
13.
IMPORTANT!!
Calculate the derivative of y = cot x.
1  cos x
tan x sin x
dy ( sin x)  sin x  cos x  cos x

dx
sin 2 x
2
2
  sin x 2cos x
sin x
 cosec2 x
y  cot x 
14.
IMPORTANT!!
y  sec x 
Calculate the derivative of y = sec x.
1  (cos x) 1
cos x
dy
 1(cos x) 2  ( sin x)
dx
 sin2x
cos x
 sin x  1
cos x cos x
 tan x  sec x
15.
IMPORTANT!!
Calculate the derivative of y = cosec x.
y  cosec x 
1  (sin x) 1
sin x
dy
 1(sin x) 2  (cos x)
dx
  cos2 x
sin x
  cos x  1
sin x sin x
  cot x  cosec x
Page 37, Exercise 5A – questions 1b, 2c, 3, 5, 7
Page 38, Exercise 6 – questions 1c, 1d, 4, 6
Page 40, Exercise 7 – questions 23
Page 7
using the Chain Rule
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
The Exponential and Logarithmic Functions
Why is e  2  718281828 ?
What is e?
Imagine there is a very generous (or stupid) banker that is offering a savings interest rate of 100%
per annum.
If I put £1 into the bank, after 1 year I will have £2.
Due to their intellect, they believe that offering 50% interest every 6 months will work out the
same.
However, now if I put £1 into the bank, after 1 year I will have £1 (1 5)2  £2  25 .
Their brilliance continues and they decide to offer an interest rate of
Now, if I put £1 into the bank, after 1 year I will now have £1 (1 
100
% payable every day.
365
1 365
)  £2  714567482 .
365
If we continue this pattern to pay interest every second then we would have
1


£1 1 

 31536000 
31536000
 £2  718281615 after 1 year.
As you can see, we are getting closer and closer to the value for e.
e is the natural growth function as it allows for ‘doubling’ with the manmade constraints of how we
measure time.
The exponential function is defined as follows,
2
3
4
e x  1  x  x  x  x  .....
2! 3! 4!
Note, that if x = 1 then
Page 8
e  1  1  1  1  1  1  1  .....
2 6 24 120 720
 2  718 (to 4 sig. fig.)
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
Derivative of ex
If
then
2
3
4
y = e x  1  x  x  x  x  .....
2! 3! 4!
dy
 0  1  x  1 x 2  1 x 3  .....
dx
2
6
2
3
 0  1  x  x  x  .....
2! 3!
x
e
i.e.
d ex  ex
 
dx
Examples:
16.
Differentiate y = e6x.
 y = eu
Let u = 6x
dy dy du


dx du dx
 eu  6
 6e6 x
17.
Calculate the derivative of y = xex.
Using the Product Rule,
Page 9
dy
 1 ex  x  ex
dx
 e x (1  x)
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
Derivative of logex (i.e. lnx)
y = logex = lnx
 x = ey
 dx  e y  x
dy
dy 1


dx x
[turn both sides upside-down]
i.e.
d ln x  1
  x
dx
Examples:
18.
Find
dy
when y = ln(5x2 – 2).
dx
 y = ln u
Let u = 5x2 – 2
dy dy du


dx du dx
 1  10 x
u
x
 10
2
5x  2
19.
Calculate the derivative of y = ln(cos x).
 y = ln w
Let w = cos x
dy
dy dw


dx dw dx
 1  ( sin x)
w
  sin x
cos x
  tan x
20.
Differentiate y = eln3x.
Let a = 3x
 y = elna
and let b = ln a
dy dy db da



dx db da dx
 eb  1  3
a
 3 eln a
a
 3 eln 3 x
3x
 1 eln 3 x
x
Page 10
 y = eb
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
Some cases can be simplified using the laws of logs (from Higher).
Differentiate y  ln 1  x2 .
1 x
2
21.
y  ln 1  x 2  ln 1  x 2   ln 1  x 2 
1 x
dy
(2 x)
 2x 2 
dx 1  x
1  x2
2 x 1  x 2   2 x 1  x 2 

1  x 2 1  x 2 
2
 4x 4
1 x
22.


Calculate d ln x 2  1 .
dx
Let y  ln x 2  1
and
u = x2 + 1
or this can simplified as:
dy dy dv du



dx dv du dx
1
 1  1 u 2  2x
v 2
 1  1  2x
u 2 u
 x
u
 2x
x 1

 y = ln v
v= u
and



Note that logs must be to the base e, otherwise, we must change the base.
For example,
y = log10x
 x = 10y
ln x = ln 10y
ln x = y ln 10
y =  ln x  1  ln x
ln10 ln10
Take log, to the base e, of both sides,
 dy 
dx
Page 11


d ln x 2  1  d 1 ln( x 2  1)
dx
dx 2
 1 d  ln( x 2  1) 
2 dx
2x

2 ( x 2  1)
1 1
ln10 x
 1
x ln10
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
Similarly, for y = 2x
Take logs of both sides to the base e,
ln y = ln 2x
ln y = x ln 2
ln y
x=
ln 2
dx  1
dy
y ln 2

dy

 y ln 2  2 x ln 2
dx
Page 43, Exercise 8A – questions 1, 3(a, c, d), 5(c, e), 6
Page 44, Exercise 8B – questions 1
Page 12
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
List of Standard Derivatives
f’(x)
xn
nxn-1
sin x
cos x
cos x
- sin x
tan x
sec2 x
cot x
- cosec2 x
sec x
sec x tan x
cosec x
- cosec x cot x
ex
ex
ln x
1
x
ax
ax ln a
loga x
1
x ln a
We don’t need
to memorise
this last two
Page 13
f(x)
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
Higher Order Derivatives – Part 1
We already know that for a function where x represents displacement and t represents time that
dx
dt
represents the change in distance with respect to the time taken for that change i.e. velocity.
d 2x
Also, 2 represents the change in velocity with respect to the time taken for that change i.e.
dt
acceleration.
Displacement, Velocity and Acceleration using Calculus
On a displacement-time graph the gradient of a chord represents average velocity whereas the
gradient of a tangentat a point represents instantaneous velocity (i.e. velocity at a certain time).
So,
v  ds
dt
distance
velocity
On a velocity-time graph the gradient of a chord represents average acceleration whereas the
gradient of a tangent at a point represents instantaneous acceleration.
So,
2
a  dv  d s2 .
dt
dt
Examples:
23.
The displacement, s metres, of a point moving in a straight line after t seconds is given by
s = 5t2 + t3.
Find the velocity and acceleration after 4 seconds.
s = 5t2 + t3
v  ds  10t  3t 2
when t = 4, v = 40 + 48 = 88
dt
 the point has a velocity of 88 m/s after 4 seconds
a  dv  10  6t
when t = 4, a = 10 + 24 = 34
dt
 the point has an acceleration of 34 m/s2 after 4 seconds.
Page 14
Bridge of Don Academy – Department of Mathematics
24.
A particle is moving in a straight line and its acceleration after t seconds is (20 – 6t) m/s2.
(a)
Find its velocity in terms of t given that its initial velocity is 5 m/s.
(b)
Find an expression for the particles displacement, s m, in terms of t given that t = 2
when s = 50.
(a)
a  dv  20  6t
dt
 v  20t  3t 2  c1
when t = 0, v = 5
(b)
Remember:
 c1 = 5
v  ds
dt
 s  5t  10t 2  t 3  c2
when t = 2, s = 50
 c2 = 8
 v = 5 + 20t – 3t2
 s = 8 + 5t +10t2 – t3
The area under the curve in a velocity-time graph gives the distance travelled.
s = area under curve =
Page 15
Mr B. A. Willox
 v dt   dt dt   ds  s  c where c is a constant
ds
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
Higher Order Derivatives – Part 2
We also use derivatives for finding stationary points or for examining the behaviour of a curve.
Here we see the graph of y  f ( x)
sloping up so
‘gradient’ is
positive
We can see that there are stationary points at
the points marked with a .
Points of inflection are marked with a .

sloping up so
‘gradient’ is
positive
sloping down
so ‘gradient’
is negative
‘horizontal’ so
‘gradient’ is zero

sloping up so
‘gradient’ is
positive

By looking at the graph of the derivative we
can see why these points must be stationary
points.
‘horizontal’ so
‘gradient’ is zero
 Here we see the graph of y  f ( x)
Notice that the point of inflection is not a
stationary point – it is an “increasing point of
inflection”.
The gradient was decreasing as in approached
the point (i.e. concave down) then after the
point the gradient was increasing (i.e. concave
up). Therefore the point is a point of
inflection.
 Here we see the graph of y  f ( x ) .
☼
☼
Page 16
Notice that the maximum stationary point from
f ( x) gives us a negative value for f ( x) and
That the minimum stationary point from f ( x)
gives us a positive value for f ( x) .
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
To summarize:
If
(i)
f (a)  0 then there is a maximum stationary point at the point  a, f (a) 
(ii)
f (a)  0 then there is a minimum stationary point at the point  a, f (a) 
(iii)
f (a)  0 then there is a point of inflection at the point  a, f (a)  .
We can use these results to determine the nature of stationary points (as an alternative to the ‘nature
table’).
However, it is often more efficient to use a ‘nature table’. For example, if we found a point of
inflection at point  a, f (a)  then we would need to use a ‘nature table’ to determine whether it was
increasing or decreasing.
Example:
25.
Sketch the curve of the function y  x 2 (2 x  3)( x  1) for 2  x  2 , indicating all roots,
stationary points and points of inflection.
2
Roots occur when y = 0  x (2 x  3)( x  1)  0
3
x  0, x   , x  1
2
y  2 x 4  x3  3x 2

dy

dx
Stationary points occur when
Page 46, Exercise 9A – questions 1, 2, 4a, 4(c, d), 6, 8
Page 46, Exercise 9B – questions 1, 3
Page 48, Review – questions 2 to 10
Page 17
dy
0
dx
Bridge of Don Academy – Department of Mathematics
Mr B. A. Willox
Examples from SQA Exams:
26.
Given that f ( x)  x e  x , x  0 obtain and simplify f ( x ) .
27.
Given f ( x)  cos 2 x e tan x , 
 
and evaluate f    .
4
Page 18

2
x

2
[2002: 4 marks]
, obtain f ( x )
[2004: 3,1 marks]
Bridge of Don Academy – Department of Mathematics
28.
29.
Page 19
Mr B. A. Willox
Obtain the derivative of each of the following functions:
(a)
f ( x)  exp (sin 2 x) ;
(b)
y  4( x
2
1)
.
Write down the derivative of tan x.
Show that 1  tan 2 x  sec 2 x .
Hence obtain  tan 2 x dx .
[2007: 3, marks]
[2008: 1,1,2 marks]
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