Mathematics for Computing
Introduction to Differentiation
Conducted by: Thilini Kulaweera
MSc (Colombo) | BSc (Kelaniya)
Module details
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Module Name: Mathematics for Computing
Module code: IT1105
Enrollment key:- IT1105
Credit Points: 4
Method of delivery:
 Lectures (2 hours/week)
 Tutorials: (1 hour/week)
2
Assessment Criteria
• Midterm -30%
 ( 1 Hour paper)
• Final Exam – 70 %

(3 hours essay-type
questions)
3
Differentiating a linear function
 A straight line has a constant gradient, or in other words, the
rate of change of y with respect to x is a constant.
 Consider the straight-line y = 3x + 2;
 Above note that y increases as a rate of 3 units, for every unit increase in x.
 We say that “the rate of change of y with respect to x is 3”.
 Observe that the gradient of the straight line is the same as the rate of change of y with
respect to x.
Key Point
 For a straight line:
“the rate of change of y with respect to x is the same as the
gradient of the line.”
Differentiation from first principles of some simple curves
 Consider the curve y = x2
 Above for different pairs of points we will get different lines, with very different
gradients.
 For a simple function like y = x2 we see that y is not changing constantly with x.
 The rate of change of y with respect to x is not a constant.
Calculating the rate of change at a point
 Calculating the rate of change at any point on a curve y = f(x) is defined to be the
gradient of the tangent drawn at that point as shown below.
 The rate of change at a point P is defined to be the gradient of the tangent at P.
Key Point
 The gradient of a curve y = f(x) at a given point is defined to be the
gradient of the tangent at that point.
 Consider the figure below which shows a fixed-point P on a curve y=x2.
 The lines through P and Q approach the tangent at P when Q is very close to P.
 Calculate the gradient of one of these lines and let the point Q approach the
point P along the curve, then the gradient of the line should approach the
gradient of the tangent at P, and hence the gradient of the curve.
 Consider a general point P which has coordinates (x, y).
 Choose point Q to be close to P on the curve.
 Because we are considering the graph of y = x2;
y  δy  (x  δx)
2
y  δy  x 2  2 x(x)  (x) 2
y  x2 ;
δy  2 x(x)  (x)
 So the gradient of PQ is;
2
δy 2 x(x)  (x) 2

x
x
δy x(2 x  x)

x
x
δy
 2 x  x
x
 As we let  x become zero, we are left with just 2x, and this is the formula for the
gradient of the tangent at P.
 Gradient of tangent =
y
lim x  lim (2 x  x)  2 x
x 0
x 0
 We can do this calculation in the same way for lots of curves. We have a special
symbol for the phrase
y
lim x
x 0
 This is again written as “dy/dx” and referred to as “derivative of y with respect to x”.
Key Point
 When n is a positive integer;
If y = xn
then
dy
n 1
 nx
dx
 The result is true when n is a negative integer and when n is a fraction
although we will not prove this here.
Linearity rules
 Also, if y = k f(x)
then,
where k is a constant
dy
df
k
dx
dx
This means that we can differentiate a constant
multiple of a function, simply by differentiating
the function and multiplying by the constant.
Linearity rules
 If
then,
y = f(x) ± g(x)
dy df dg


dx dx dx
This means that we can differentiate sums (and differences) of
functions, term by term.
The Chain Rule
 The chain rule, exists for differentiating a function of another
function.
 Consider the expression (x4+x2-9)10. We can call such an
expression a ‘function of a function’.
 Suppose, in general, that we have two functions, f(x) and g(x).
Then y = f(g(x)) is a function of a function.
 g(x) = x4+x2-9 and f(x) = x10
f(g(x)) = f(x4+x2-9) = (x4+x2-9)10
Key Point
 To differentiate y = f(g(x)), let u = g(x).
Then y = f(u) and
dy dy du


dx du dx
Exercise
 Differentiate
y = (2x-5)10
Let u = 2x-5 so that y = u10. It follows that
dy dy du


dx du dx
9
 10u  2
 20(2 x  5) 9
du
2
dx
dy
 10u 9
du
The Product Rule
 The product rule: if y = uv then
dy
dv
du
u v
dx
dx
dx
Exercise
 Find the derivative of y = (3x – 2x2) (5 +4x).
dy
d
2 d
 (3x  2 x ) [5  4 x]  (5  4 x) [3x  2 x 2 ]
dx
dx
dx
 (3 x  2 x 2 )(4)  (5  4 x)(3  4 x)
 (12 x  8 x )  (15  8 x  16 x )
2
 15  4 x  24 x
2
2
The Quotient Rule
 The quotient rule: if y = u then
v
du
dv
v
u
dy
dx
dx

2
dx
v
Exercise
 Find the derivative of
x 1
y
2x  3
d
d
(2 x  3)
[ x  1]  ( x  1)
(2 x  3)
dy
dx
dx

2
dx
(2 x  3)
(2x  3)(1)  (x  1)(2)

2
(2x  3)
5

2
(2x  3)
Higher- order derivatives
The derivative of f’ is the second derivative of f and is denoted by f .
d
[ f ' ( x)]  f " ( x)
dx
The derivative of f is the third derivative of f and is denoted by f .
Example
f ( x)  2 x  3 x
4
2
f ' ( x)  8 x 3  6 x
f ( x)  24 x  6
"
2
f ' " ( x)  48 x
f
( 4)
( x)  48
f
(5)
( x)  0
Explicit Vs Implicit
 Explicit is when the function shows us how to go
directly from x to y, such as;
y = x3 + 3
(That is the classic y = f(x) style)
 Implicit is when it is not given directly such as;
x2 − 3xy + y3 = 0
27
Implicit differentiation
 Sometimes functions are given not in the form y = f(x) but in a more complicated form in
which it is difficult or impossible to express y explicitly in terms of x.
 Such functions are called implicit functions.
 Now we look at how we might differentiate functions of y with respect to x.
 Consider an expression such as;
x2 + y2 − 4x + 5y − 8 = 0
 It would be quite difficult to re-arrange this so y was given explicitly as a
function of x.
Key Point
d
d
dy
( f ( y ))  ( f ( y )) 
dx
dy
dx
Remember, every time we want to differentiate a function of y with
respect to x, we differentiate with respect to y and then multiply by
(dy/dx).
Example We differentiate each
term with respect to x:
rearrange
Suppose we want to differentiate the implicit function
with respect to x.
y 2  x3  y 3  6  3 y
d 2
d 3
d 3
d
d
( y )  ( x )  ( y )  (6)  (3 y )
dx
dx
dx
dx
dx
d 2 dy
d
dy
d
dy
( y )   3 x 2  ( y 3 )   0  (3 y ) 
dy
dx
dy
dx
dy
dx
dy
dy
2
2 dy
2 y  3x  3 y
3
dx
dx
dx
dy
dy
dy
3x 2  3 y 2
 2y  3
dx
dx
dx
dy
3 x 2  (3 y 2  2 y  3)
dx
dy
3x 2

dx (3 y 2  2 y  3)
Relationship of Exponential and Logarithmic
“The natural Logarithmic function and the natural exponential
function are inverses of each other.”
Since the exponential function is the inverse of logarithmic
function;
Domain of the exponential function = Range of the logarithmic function =
Set of all real numbers
Range of the exponential function = Domain of the logarithmic function =
(0,+∞)
Properties of Exponents
 Let a and b be positive numbers
1. 𝑎0 = 1
2. 𝑎 𝑥 𝑎 𝑦 = 𝑎 𝑥+𝑦
3.
𝑎𝑥
𝑎𝑦
=𝑎
𝑥−𝑦
4. (𝑎 𝑥 )𝑦 = 𝑎 𝑥𝑦
5. (𝑎𝑏)𝑥 = 𝑎 𝑥 𝑏 𝑥
6.
𝑎 𝑥
𝑏
7. 𝑎
−𝑥
=
=
𝑎𝑥
𝑏𝑥
1
𝑎𝑥
Properties of Logarithms
 Let U and V be real positive numbers:
1. ln 1 = 0
2. ln 𝑈𝑉 = ln 𝑈 + ln 𝑉
3. ln
𝑈
𝑉
= ln 𝑈 − ln 𝑉
4. ln 𝑈 𝑥 = 𝑥. 𝑙𝑛 𝑈
5. ln 𝑒 𝑥 = 𝑥
6. 𝑒 ln 𝑥 = 𝑥, 𝑥 > 0
7. ln 𝑈 = ln 𝑉 𝑖𝑠 equivalent 𝑡𝑜 𝑈 = 𝑉
Task 1
 Expand the Logarithmic expression given (Assume x > 0 and y> 0):
1.
10
ln
9
2. ln 𝑥 2 + 1
3.
𝑥𝑦
ln
5
4.
𝑥2
ln 3
6𝑦
Task 2
 Condense the Logarithmic expressions given:
1. 3 ln 𝑈 − ln(𝑈𝑉)3 +3 ln(𝑈𝑉)
2. ln 𝑈 2 − 𝑉 2 − ln 𝑈 − 𝑉 − ln(𝑈 + 𝑉)
Task 3
 Solve the following equations for the variable.
1. 𝑒 𝑥 = 5
2. 10 + 3𝑒 0.1𝑡 = 14
3. ln 𝑥 = 5
4. 3 + 2 ln 𝑥 2 = 7
Derivatives of Natural Exponential Functions
 If 𝑦 = 𝑒 𝑥 ,
𝑑(𝑒 𝑥 )
𝑑𝑥
= 𝑒𝑥
 If 𝑦 = 𝑒 𝑓(𝑥) ,
𝑑(𝑒 𝑓 𝑥 )
𝑑𝑥
= 𝑒𝑓
𝑥
.𝑓′(𝑥)
 These are rules that we need to remember - don't worry about where
they come from.
Example 1:
 Find the derivative of 𝑦 = 𝑒
5𝑥 2
.
We let f 𝑥 = 5𝑥 2 . Then 𝑓 ′ 𝑥 = 10𝑥, and
2
𝑑(𝑒 5𝑥 )
𝑑𝑥
= 10𝑥𝑒
5𝑥 2
Now it’s your turn!
Differentiate the given exponential functions.
1. 𝑓 𝑥 = 𝑒 3𝑥
2. 𝑓 𝑥 = 𝑒 6−𝑥
3. 𝑓 𝑥 = 𝑒 −𝑥
4. 𝑦 = −3𝑒 −3𝑥
5. 𝑦 = 𝑥𝑒 𝑥
𝑒𝑥
𝑥
6. 𝑓 𝑥 =
7. 𝑓 𝑥 = 𝑥𝑒 𝑥 − 𝑒 𝑥
1 𝑥2 +2𝑥−1
8. 𝑓 𝑥 = −3𝑒
Derivatives of Natural Logarithmic Function
 If 𝑦 = ln 𝑥,
𝑑 ln 𝑥
𝑑𝑥
=
1
𝑥
=
1
𝑓(𝑥)
 If 𝑦 = ln 𝑓(𝑥),
𝑑(ln 𝑓(𝑥))
𝑑𝑥
× 𝑓′(𝑥)
Example 2:
 Find the derivative of the function 𝑦 = ln(6𝑥 + 3).
Let 𝑓 𝑥 = 6𝑥 + 3.Then 𝑓 ′ 𝑥 = 6 so that,
𝑑(ln(6𝑥+3))
6
=
𝑑𝑥
6𝑥+3
Now it’s your turn!
Differentiate the given logarithmic functions.
1. 𝑓 𝑥 = ln 2𝑥
2. 𝑓 𝑥 = ln(2𝑥 2 + 4)
3. 𝑦 = 𝑥𝑙𝑛 𝑥
4. 𝑦 = ln 𝑥 + 1
5. 𝑓 𝑥 = ln
𝑥(𝑥 2 +1)2
2𝑥 3 +1
Thank You !!