THE DERIVATIVE AND DIFFERENTIATION
OF ALGEBRAIC FUNCTIONS
OBJECTIVES:
• to define the derivative of a function
• to find the derivative of a function by
increment method (4-step rule)
• to identify the different rules of differentiation
and distinguish one from the other;
• prove the different rules of differentiation using
the increment method;
• find the derivative of an algebraic function using
the basic rules of differentiation; and
• extend these basic rules to other “complex”
algebraic functions.
Derivative of a Function
The process of finding the derivative of a
function is called differentiation and the
branch of calculus that deals with this
process is called differential calculus.
Differentiation is an important mathematical
tool in physics, mechanics, economics and
many other disciplines that involve change
and motion.
Consider a point Q( x2 , f ( x2 )) on the curve
y  f (x), that is distinct from P( x1 , f ( x1 )),and
compute the slope mPQ of the secant line
through P and Q.
mPQ
f ( x2 )  f ( x1 )

x
m PQ
where : x  x2  x1
and x2  x1  x
f ( x1  x)  f ( x1 )

x
If we let x2 approach x1, then the point Q
will move along the curve and approach
point P. As point Q approaches P, the value
of Δx approaches zero and the secant line
through P and Q approaches a limiting
position, then we will consider that position
to be the position of the tangent line at P.
y
tangent line
P( x1 , f ( x1 ))
Q( x2 , f ( x2 ))
secant line
y
y  f (x)
x
x  x 2  x1
x 2  x1  x
Thus, we make the following definition.
DEFINITION:
Suppose that x1 is in the domain of the function
f, the tangent line to the curve y=f(x) at the point
P(x1,f(x1)) is the line with equation,
y  f ( x1 )  m( x  x1 )
f ( x1  x )  f ( x1 )
where m  lim
provided
x 0
x
the limit exists, and P( x1 , f ( x1 )) is the point
of tangency.
DEFINITION
The derivative of y = f(x) at point P on the
curve is equal to the slope of the tangent line at
P, thus the derivative of the function f given by
y= f(x) with respect to x at any x in its domain is
defined as:
dy
y
f ( x  x)  f ( x)
 lim
 lim
dx x0 x x0
x
provided the limit exists.
Other notations for the derivative of a function are:
d
D x y, D x f ( x), y ' , f ' , f ' ( x), and
f ( x)
dx
Note:
To find the slope of the tangent line to the curve at point
P means that we are to find the value of the derivative at
that point P.
There are two ways of finding the derivative of a function:
1. By using the increment method or the four-step rule
2. By using the differentiation formulas
THE INCREMENT METHOD OR THE FOUR-STEP RULE
One method of determining the derivative of
a function is the increment method or more
commonly known as the four-step rule.
The procedure is as follows
:
STEP 1: Substitute x + Δx for x and
y + Δy for y in y = f(x)
STEP 2: Subtract y = f(x) from the result of
step 1 to obtain Δy in terms of x
and Δx
STEP 3: Divide both sides of step 2 by Δx.
STEP 4: Find the limit of step 3 as Δx
approaches 0.
EXAMPLE
dy
1. Find
u sin g the four  step rule given y  1  x 2 .
dx
2
y
a. y  y  1  x  x
d . lim
 lim  2 x  x 
x 0 x
x 0
2
b. y  y  y  1  x  x   y
dy
2
 2 x
2
y  1  x  2 xx  x  y
dx
y  1  x  2 xx  x  1  x 2 
2
2
2
y  1  x  2 xx  x  1  x 2
2
y  2 xx  x
2
y x 2 x  x 
c.

x
x
EXAMPLE 2. Find dy u sin g the four  step rule given y  2 x  1
dx
a. y  y 
1  2x
2( x  x )  1
1  2( x  x )
2( x  x )  1 2 x  1

1  2( x  x ) 1  2 x
2 x  2x  11  2 x   2 x  11  2 x  2x 
y 
1  2 x  2x 1  2 x )
b. y 
2 x  2x  1  4 x 2  4 xx  2 x  2 x  4 x 2  4 xx  1  2 x  2x
y 
1  2 x  2x 1  2 x )
c.
y
4 x

x x( 1  2 x  2x )( 1  2 x )
y
4
 lim
x 0 x
x 0 ( 1  2 x  2 x )( 1  2 x )
d . lim

dy
4

dx ( 1  2 x )2
EXAMPLE
dy
3. Find
u sin g the four  step rule given y  x  1 when x  10
dx
a. y  y  x  x  1
b. y 
when x  10,
dy
1
1
1



dx 2 10  1 2( 3 ) 6
x  x  1  x  1
y
x  x  1  x  1

x
x
y
x  x  1  x  1
x  x  1  x  1
d. lim
 lim

x  0  x
x 0
x
x  x  1  x  1
x  x  1  x  1
 lim
x 0 x
x  x  1  x  1
x
 lim
x 0 x
x  x  1  x  1
y
1
lim
 lim
x  0  x
x 0
x  x  1  x  1
c.





dy
1

dx 2 x  1
DIFFERENTIATION OF
ALGEBRAIC FUNCTIONS
DERIVATIVE USING FORMULAS
The increment-method (four-step rule) of
finding the derivative of a function gives us the
basic procedures of differentiation. However
these rules are laborious and tedious when the
functions to be differentiated are “complex”,
that is, functions with large exponents,
functions with fractional exponents and other
rational functions
Understanding of the theorems of
differentiation is very important. This is the
heart of differential calculus. All of the
succeeding topics such as applications of
derivatives, differentiation of transcendental
functions etc. will be dependent on these
theorems. Understanding of these theorems
will enable us to calculate derivatives more
efficiently and will make calculus easy and
enjoyable.
DIFFERENTIATION FORMULAS
Derivative of a Constant
Theorem: The derivative of a constant function
is 0; that is, if c is any real number,
then,
d
[c ]  0
dx
Exam ple: Differentiate the following functions.
1. y  5
y'  0
2. y   25
dy
0
dx
3
3. f(x)  4
f' (x)  0
4. h(x)  log3 4
h' (x)  0
DIFFERENTIATION FORMULAS
Derivatives of Power Functions
Theorem: ( Power Rule) If n is a positive integer,
then,
d n
n 1
[ x ]  nx
dx
In words, to differentiate a power
function, decrease the constant exponent
by one and multiply the resulting power
function by the original exponent
.
Exam ple: Differentiate the following functions
3. f(x)  x 8
1. y  x 4
y'  4 x
4 1
y'  4 x
3
2. y  x
6
7
f' (x)  -8 x
9
8
 9
x
4. F(x)  x log3 4
6
dy 6 7 1
 x
dx 7
dy 6
 x
dx 7
f' (x)  -8 x
 8 1
6 7

7 7
F' (x)   log3 4 x 
 log3 4 1
6
 x
7

1
7
6 7 x6
 7 
7x
7 x
6
DIFFERENTIATION FORMULAS
Derivative of a Constant Times a Function
Theorem: ( Constant Multiple Rule) If f is a
differentiable function at x and
c is any real number, then cf is
also differentiable at x and
d
d
cf ( x)  c  f ( x)
dx
dx
In words, the derivative of a constant
times a function is the constant times the
derivative of the function, if this derivative
exists.
Proof:
d
cf ( x  x)  cf ( x)
cf ( x)  lim
x 0
dx
x
 f ( x  x)  f ( x) 
 lim c 

x 0
x


f ( x  x)  f ( x)
 c lim
x 0
x
d
 c  f (x)
dx
Exam ple: Differentiate the following functions
3. f(x)  9 x 4
1. y  5 x 8
y'  5 8 x
f' (x)   9 - 4 x
36
5
f' (x)  36 x  5
x
4 3
4. F(r)  r
3
4 
2
F' (r)    3r   4 r 2
3 
 4 1
7
y'  40x7
2. y  5 x
2
5
dy
2
  5  x
dx
5
dy
 2 x
dx
2 5

5 5
2
1
5
 2 x

3
5

2
5
2 x

x
x3
5
2
DIFFERENTIATION FORMULAS
Derivatives of Sums or Differences
Theorem: ( Sum or Difference Rule) If f and g are
both differentiable functions at x,
then so are f + g and f – g, and
d
d
d
 f  g    f   g 
dx
dx
dx
or
d 
d
d

f ( x)  g ( x)   f ( x)  g ( x)

dx 
dx
dx

In words, the derivative of a sum or of a
difference equals the sum or difference of
their derivatives, if these derivatives exist.
Proof:
d
[ f ( x  x)  g ( x  x)  [ f ( x)  g ( x)]
[ f ( x)  g ( x)]  lim
x 0
dx
x
[ f ( x  x)  f ( x)]  [ g ( x  x)  g ( x)]
 lim
x 0
x
f ( x  x)  f ( x)
g ( x  x)  g ( x)
 lim
 lim
x 0
x 0
x
x

d
d
[ f ( x)]  [ g ( x)]
dx
dx
Exam ple: Differentiate the following functions
3. f(x)  2 x 4  9 x  4
1. y  5 x 4  6 x 2  4x  7
f' (x)  8 x 5  9
-8
f' (x)  5  9
x
y'  20x 3  12x  4


y'  4 5x3  3 x  1
4
3
2
2. y  6 x  2 x  4x  5 x  9
2
1
dy
15
 24x 5 - 4x - 4 - x 2
dx
2
dy 24
15 21
 5 - 4x - 4 - x
dx x
2
4 3
4. F(r)  r  r
3
4 
2
F' (r)  - 2r -3    3r 
3 
2
- 2
F' (r)  3  4r 2
r
DIFFERENTIATION FORMULAS
Derivative of a Product
Theorem: (The Product Rule) If f and g are both
differentiable functions at x, then so
is the product f ● g, and
d
dg
df
or
 f  g  f  g
dx
dx
dx
d
d
d
 f ( x)  g ( x)  f ( x) [ g ( x)]  g ( x)  f ( x)
dx
dx
dx
In words the derivative of a product of two
functions is the first function times the derivative
of the second plus the second function times the
derivative of the first, if these derivatives exist.
Proof:
d
f ( x  x)  g ( x  x)  f ( x)  g ( x)
[ f ( x)  g ( x)]  lim
x 0
dx
x
 lim
x 0
f ( x  x)  g ( x  x)  f ( x  x)  g ( x)  f ( x  x)  g ( x)  f ( x)  g ( x)]
x
g ( x  x)  g ( x)
f ( x  x)  f ( x) 

 lim  f ( x  x) 
 g ( x) 

x0
x
x

g ( x  x)  g ( x)
f ( x  x)  f ( x)
 lim g ( x)  lim
x 0
x 0
x 0
x
x
 lim f ( x  x)  lim
x 0

 lim f ( x  x
x 0
dxd g ( x)  lim g ( x)dxd  f ( x)
x 0
Exam ple: Differentiate the following functions and sim plify.


3 x  4 8 x   4 x
1. y  3 x  4  4 x 2  3
y' 
2

 3 3
y'  24x2  32x  12x2 - 9
y'  36x2  32x- 9

y'  x

 1- 2   5 - 2x3x 
2. y  x 3  1 5 - 2x
3
y'  -2x3  2  15x2 - 6x3
y'  -8x3  15x2  2
2
3. y  2  x 3  10x  5 
y'  2  x 3  10   10x  5  3 x 2 
y'  -20  10x 3  30x 3  15x 2
y'  40x 3  15x 2 - 20
y'  58x3  3 x 2  4 
DIFFERENTIATION FORMULAS
Derivative of a Quotient
Theorem: (The Quotient Rule) If f and g are both
differentiable functions at x, and if
g(x) ≠ 0 then f/g is differentiable at x
and
df
dg
g
f
d f
or
   dx 2 dx
dx  g 
g
d
d
g ( x)  f ( x)  f ( x) g ( x)
d  f ( x) 
dx
dx



dx  g ( x) 
g ( x)2
In words, the derivative of a quotient of
two functions is the fraction whose
numerator is the denominator times the
derivative of the numerator minus the
numerator times the derivative of the
denominator and whose denominator is
the square of the given denominator
Note:
The proof is left for the students .
Exam ple: Differentiate the following functions and sim plify.
4 x2  3
1. y 
1  2x

1  2 x 8 x   4 x 2  3  2 
y' 
2
1  2 x 


8 x  16 x 2  8 x 2  6
y' 
2
1  2 x 
y' 
 8 x2  8 x  6
1  2 x 2
y' 
2  4 x2  4 x  3
1  2 x 2


DIFFERENTIATION FORMULAS
Derivatives of Composition
Theorem: (The Chain Rule) If g is differentiable at
x and if f is differentiable at g(x), then
the composition f ◦ g is differentiable
at x. Moreover, if y=f(g(x)) and u=g(x)
then y=f(u) and
dy dy du


dx du dx
or
 
d un
n 1 du
 nu
dx
dx
Exam ple: Differentiate the following functions and sim plify.


1. y  3x  10x  15
2

5

y'  5 3x  10x  15 6x - 10
4
2
 5 
2. G(x)  

 x -1
3
 5 
G' (x)  3

 x -1
2

 125 
5 
 
  3

2 
4 
  x  1 
  x  1 
3. y  3x  14 x  5 
4
y  43x  14 x  5  3x  14   4 x  5 3 
3
y  43x  14 x  5  12x  4  12x - 15
3
y  424x - 11 3x  14 x  5 
3
DIFFERENTIATION FORMULAS
Derivative of a Radical with index equal to 2
If u is a differentiable function of x, then
du
d
dx
u 
dx
2 u
 
The derivative of a radical whose index
is two, is a fraction whose numerator is
the derivative of the radicand, and whose
denominator is twice the given radical, if
the derivative exists.
DIFFERENTIATION FORMULAS
Derivative of a Radical with index other than 2
If n is any positive integer and u is a
differentiable function of x, then
1
1


d
1 n 1 du
n
u    u 
dx   n
dx
The derivative of the nth root of a given
function is the exponent multiplied by the
product of u whose power is diminished by
one and the derivative of u, if this derivative
exists.
Differenti ate the following and sim plify.
1. H  x   3 x  5
3
3x  5 3 3x  5
H'  x  


2 3x  5
3 x  5 23 x  5 
2. y  5 2 x  4  x  5 
y  2 x  4  x  5 
1
5
1
1
1
y'  2 x  4  x  5 5 2 x  4 1   x  5 2 
5
4
1

y'  2 x  4  x  5  5 2 x  4  2 x  10
5
4
1

y'  2 x  4  x  5  5 4 x  14
5
4
1

y'  2 x  4 x  5  5 2 x  4  2 x  10
5
4
1

y'  2 x  4 x  5  5 4 x  14
5
22 x  7 
y' 
4
52 x  4 x  5  5
EXERCISES
I. Find the derivative of the following functions
using the four-step rule
.
1. f ( x )  4 x  5 x
2
2. f ( x )  x
1
2
2
6. y  ax  b
2


7. y  x  3 x  3 x  1
3
2
2
3. f ( x ) 
x1
2x  3
8. y 
1  2x
x1
4. y 
x
9. y  ( 2  x )( x  3 )
5. y  4 x  3
10. y  1  x 3
1
3
EXERCISES
II. Find the derivative of the following functions
using the four-step rule
.
ds
1. Given s  t  2 , find
dt
dA
2
2. Given A  r , find
dr
4 3
dV
3. Given V  r , find
3
dr
2t  3
dS
4.Given S 
, find
3t  4
dr
EXERCISES
III. Evaluate the derivative of the following functions
using the four-step rule with the indicated value
of x.
1. y 
1
when x  8.
2x
x-2
2. y 
when x  6.
x-3
a2
3. y 
when x  a
ax
4. y  x  12 x  3 when x  5
5. y  x 3  1 when x  3 4
6. If f ( x )  x 3  3 x 2 ,
find the valuesof x
for which f ' ( x )  0