Workshop: Differentiation 1
Introduction to Differentiation
Topics Covered:
• Differentiation of elementary functions.
• Linearity rule of differentiation.
• Higher order derivatives.
• Differentiation applications: stationary points and point of inflexion
by Dr.I.Namestnikova
1
Differentiation of Elementary Functions
The derivative of a real function y
= f (x) is usually denoted by one of these
symbols
dy
dx
d
dx
= f ′ (x) =
d
dx
[f (x)] ≡ f (1) (x)
is called a differential operator and x is a differentiation variable.
To calculate the derivative of an elementary function the differential table can
be used.
function to differentiate
result of differentiation
f (x)
f ′ (x)
C , constant
0
xn
nxn−1
ln x
1
x
sin x
cos x
cos x
sin x
tan x
sec x
arcsin x = sin−1 x
√1
arccos x = cos−1 x
√−1
1−x2
1−x2
1
1+x2
arctan x = tan−1 x
Note if a function not depend on the variable of differentiation, then it should be
differentiate as a constant. For example
2
d
dx
[t2 − 3t + 4] = 0.
Example 1
Differentiate y
= x3 .
Solution.
To get the result we can use the formula from the differentiation table
putting n
= 3. Therefore we have,
d
[|{z}
x3 ] = 3 × x3−1 = 3x2
| {z }
dx
xn
Example 2
Differentiate y
n×xn−1
= tan x.
Solution.
According to the differentiation table we obtain
d
dx
[tan x] = sec x
Linearity rule of Differentiation
d
•
•
dx
d
dx
[αf (x)] = α
d
dx
[f (x)±g(x)] =
[f (x)] - constant factor α out
d
dx
[f (x)]±
d
dx
[g(x)] - separate terms
or
d
dx
[αf (x)±βg(x)] = α
where α and β are constants.
3
d
dx
[f (x)]±β
d
dx
[g(x)]
Example 3
Differentiate y
= 7x4 + 9x2 .
Solution.
Using the linearity rule we obtain
d
4
d
2
d
4
2
[7x
+ 9x}] = 7 [|{z}
x ] + 9 [|{z}
x ]
|
{z
dx
dx xn
dx xn
α=7
β=9
f (x)=x4
g(x)=x2
n=4
n=2
= 7 × 4x4−1 + 9 × 2x2−1
= 28x3 + 18x
separate terms
and factor out
use the differentiation
table
Example 4
Differentiate y
Solution.
= 8 sin x + 2x5 − 3 ln x.
Using the linearity rule for three terms we have
d
[8 sin x + 2x5 − 3 ln x]
dx
d
d 5
d
separate terms
= 8 [sin x] + 2 [x ] − 3 [ln x]
and factor out
dx
dx
dx
1
use the differentiation
5−1
= 8 cos x + 2 × 5x
−3
table
x
3
= 8 cos x + 10x4 −
x
Example 5
Find
dy(t)
dx
if y(t)
= 9 cot t − 7t3 .
Solution.
dy(t)
We get dx = 0 because the function y(t) is a function of variable t
and does not depend on differentiation variable x.
4
Excises
1. Differentiate with respect to x
f (x) = 5x + 9x2
√
2
(b) f (x) = 2 x − 8x− 3
(a)
f (x) = 9 tan(x) − 7x5
1
(d) f (x) =
cos(x) − x−3
30
2
−2
− 5 sin x
(e) f (x) = − ln(x) + 7x
3
2. Find d
1
(a)
3t4 − et + 9
dt
5
d
1
(b)
3t4 − et + 9
dy
5
1
d
− 2ez + 5 sin z
(c)
2
dz z
d 3
−2
+ 24 tan y
13 ln y − 9y
(d)
dy
(c)
Answers
1. (a)
(b)
(c)
(d)
(e)
2. (a)
(b)
(c)
(d)
f ′ (x) = 5 + 18x
2
16 − 5
x 3
f ′ (x) = − √ +
x
3
f ′ (x) = 9 sec2 (x) − 35x4
1
f ′ (x) = −
sin(x) + 3x−4
30
2
f ′ (x) = −
− 14x−3 − 5 cos x
3x
1
12t3 − et
5
0 (the function does not depend on y )
2
− 3 − 2ez + 5 cos z
z
13
27 − 5
+
y 2 + 24 sec2 y
y
2
5
Higher Order Derivatives
f ′ (x) is itself also a real function. Therefore
The derivative of a real function
we can differentiate it once more, e. g.,
d
df
dx dx
=
d
dx
′
f (x) ≡
d
dx
f (1) (x)
As a result we have a second order derivative which we denote as
d2 f
dx2
≡ f ′′ (x) ≡ f (2) (x)
Now we can define the n-th order derivative,
f
(n)
(x) =
d dx
f
(n−1)
(x)
Note: one should calculate the first order derivative before to calculating the
second order derivative.
Example 6
Find f (2) (x) if f (x)
= sin x.
Solution.
Using the differentiation table we obtain first
d
dx
[sin x] = cos x
By differentiation the result we get the required derivative
d2
d
[sin
x]
=
[cos x] = − sin x
dx2
dx
6
Example 7
Find f (2) (x) if f (x)
Solution.
= 3x5 − 7x4 + 4 ln x.
Using the linearity rule and the differentiation table we obtain first
d
[3x5 − 7x4 + 4 ln x]
dx
d 5
d 4
d
separate terms
= 3 [x ] − 7 [x ] + 4 [ln x]
and factor out
dx
dx
dx
1
use the differentiation
5−1
4−1
= 3 × 5x
− 7 × 4x
+4
table
x
= 15x4 − 28x3 + 4x−1
By differentiation the above result we get the required second order derivative
d2
d
5
4
[15x4 − 28x3 + 4x−1 ]
[3x
−
7x
+
4
ln
x]
=
2
dx
dx
= 60x3 − 84x2 − 4x−2
Example 8
Find f (4) (x) if f (x)
= 7x−2 + 3ex + 9.
Solution. Using the linearity rule and the differentiation table we obtain first
d
dx
[7x−2 + 3ex + 9] = 7
Further we have
f (2) (x) =
f (3) (x) =
f (4) (x) =
d
[x−2 ] + 3
d
dx
dx
= −14x−3 + 3ex
d
dx
d
dx
d
dx
[ex ] + 0
[−14x−3 + 3ex ] = 52x−4 + 3ex
[52x−4 + 3ex ] = −208x−5 + 3ex
[−208x−5 + 3ex ] = 1040x−6 + 3ex
7
Excises
Find f (2) (x) and f (3) (x) if
f (x) = 5x + 9x2
√
2
2. f (x) = 2 x − 8x− 3
1.
f (x) = 9 ln(x) − 3x5
1
cos(x) − x−3
4. f (x) =
30
2
−2
5. f (x) = − ln(x) + 7x
− 5 sin x
3
3.
Answers
1.
f (2) (x) = 18,
f (3) (x) = 0.
2.
1 3
80 − 8
f (2) (x) = − x− 2 −
x 3
2
9
640 − 11
3 −5
x 2+
x 3
f (3) (x) =
4
27
3.
f (2) (x) = −9x−2 − 60x3
f (3) (x) = 18x−3 − 180x2
4.
f (2) (x) = −12x−5 −
f (3) (x) = 60x−6 +
1
30
1
30
cos x
sin x
5.
f (2) (x) = 42x−4 +
2
3
x−2 + 5 sin x
f (3) (x) = −168x−5 −
8
4
3
x−3 + 5 cos x
Stationary points and point of inflexion
y = f (x), between two variable quantities x and y
can be represented by a graph of y against x. Any point (x0 , y0 ) on
dy
Any relationship,
the graph at which
dx
takes value zero is called a stationary point.The
tangent to the curve at the stationary point (x0 , y0 ) is parallel to x-axis.
5
2.5
f HxL
Stationary point
0
-2.5
Stationary point
-5
-7.5
0
-1
1
2
3
x
4
5
2
A stationary point
1
Point of Inflexion
f HxL
(x0 , y0 )
on the graph is said to be a
0
point of inflexion if the curve
-1
exhibits a change in the bend-2
ing direction there.
-1.5
-1
-0.5
0
x
0.5
1
1.5
9
2
Local Maxima
5
2.5
f HxL
Local Maximum
0
-2.5
-5
-7.5
0
-1
1
2
3
x
4
5
A stationary point (x0 , y0 ) on the graph is said to be a local maximum
if y0 is greater than the y coordinates of all other points on the curve in
the immediate neighbourhood of (x0 , y0 ).
Note: It may well happen that, for points on the curve at some distance
away from a local maximum
(x0 , y0 ), their y coordinates are greater
y0 . Hence, the definitions of a local maximum point must refer to
the behaviour of y in the immediate neighbourhood of the point.
then
Local Minima
A stationary point
5
(x0 , y0 )
on the graph is said to be
2.5
f HxL
a local minimum if
0
y0 is
less than the y coordinates of
-2.5
Local Minimum
-5
all other points on the curve
-7.5
-1
0
1
2
x
3
4
5
in the immediate neighbourhood of (x0 , y0 ).
10
The Location of Stationary Points and their Nature
First, we solve the equation
dy
= 0. Having located a stationary point (x0 , y0 ),
dx
we then determine whether it is local maximum or local minimum point.
Method 1. The First Derivative Method.
For x in the immediate neighbourhood of the local minimum point x0
x < x0
x = x0
x > x0
dy
dx
negative
0
positive
Direction
ց
−→
ր
of graph
For x in the immediate neighbourhood of the local maximum point x0
x < x0
x = x0
x > x0
dy
dx
positive
0
negative
Direction
ր
−→
ց
of graph
For x in the immediate neighbourhood of the horizontal point of inflection x0
x < x0
x = x0
x > x0
dy
dx
negative
0
negative
Direction
ց
−→
ց
of graph
or
x < x0
x = x0
x > x0
dy
dx
positive
0
positive
Direction
ր
−→
ր
of graph
11
Example 9
Find stationary points of the function y
= x3 − 6x2 + 9x − 3.
Solution.
Stationary points:
dy
dx
dy
⇒
dx
x<1
dy
dx
dy
|
dx x=0
= 3x2 − 12x + 9
= 0 at x = 1,
x=1
=9
1<x<3
dy
|
dx x=2
0
positive
x=3
= −3
0
negative
ր
Direction
x = 3.
−→
ց
−→
⇒ when x = 1, y = 1 - local maximum
when x = 3, y = −3 - local minimum
f HxL
5
Local Maximum
x
1
-1
2
3
4
-2.5
Local Minimum
-5
-7.5
12
dy
|
dx x=4
=9
positive
of graph
2.5
x>3
5
ր
Method 2.
The Second Derivative Method.
dy
|
dx x=x0
d2 y
|
dx2 x=x0
0
0
negative
positive
T
S
Graph
local max
local min
Example 10
Find stationary points of the function f (x)
Solution.
= x3 − 6x2 + 9x − 3.
Stationary points:
df
dx
≡ 3x2 − 12x + 9 = 0 at x = 1,
d2 f
dx2
df
dx
d2 f
dx2
= 6x − 12
x=1
x=3
0
0
d2 f
|
dx2 x=1
= −6
d2 f
|
dx2 x=3
=6
negative
positive
T
S
Graph
local max
local min
f HxL
5
2.5
Local Maximum
x
1
-1
2
3
4
5
-2.5
Local Minimum
-5
-7.5
13
x = 3.
Example 11
1. Obtain the critical points and determine their nature for the function
f (x) = (x − 1)(x − 2)(x − 3) = x3 − 6x2 + 11x − 6
2. Sketch the function graph highlighting key points on the curve.
Solution.
1. Find zeros of function f (x)
f (x) = 0 at poins x = 1,
x = 2,
x=3
2. Find Stationary points:
df
= 3x2 −12x+11,
dx
df
= 0 at x = 1.4227,
dx
d2 f
dx2
= 6x − 12
x = 1.4227
x = 2.5774
f (x)
0.3849
df
dx
d2 f
dx2
−0.3849
0
0
d2 f
|
dx2 x=1.4227
Graph
= −3.4641
d2 f
|
dx2 x=2.5774
= 3.4641
negative
positive
T
S
local max
local min
fHxL
1.5
1
0.5
1
2
x = 2.5774
3
4
-0.5
-1
-1.5
14
x
Example 12
Find stationary points of the function f (x)
= 3x3 .
Solution.
Stationary points:
df
dx
= 9x2 ,
⇒
df
dx
= 0 at x = 0.
x<0
x = x0
x>0
dy
dx
positive
0
positive
Direction
ր
−→
ր
of graph
Therefore x
= 0 is point of inflexion of the curve f (x) = 3x3 .
2
1.5
1
0.5
-2
-1
1
-0.5
-1
-1.5
-2
2
fHxL = 3 x3
15