LEARNING MODULE No.

2
TOPIC: Basic Differentiation Techniques

BACKGROUND INFORMATION FOR LEARNERS:

In the previous module, the derivation of the derivative of a function is explained in

details. By applying rules of limits and illustrating through graphs of curves and lines, you
learned that the derivative of a function can be solved using the Increment Method. All
derivatives of functions can be solved using the said method, but in cases where in the
function is complicated, though possible, the process is long and tedious. A need therefore to
establish rules in solving the derivatives of functions is but necessary. Since differential
equations topics require you to master differentiation techniques, it is therefore mandatory to
have a run-through of the different techniques of differentiation. This will enable you to find
derivatives easily.

Differentiation Rules
1. The Constant Rule
If f ( x )=c and c is constant, then f ' ( x )=o
The derivative therefore of any constant function is zero.

Example:

Solve y’ of y=20

Solution:
Since y=20 is a constant function, then y '=0

2. The Power Rule

A power function can be written in the form f ( x )=x n , where n is real number.
If f ( x )=x n , its derivative f ' ( x )=n x n−1

Example:
What is the derivative of the function f ( x )=x 12
 Solution:
Since the function f ( x )=x 12 is a Power Function then, f ' ( x )=12 x 11

3. The Constant Multiple Rule

If f ( x )=cg( x ), then f ' ( x )=c g' (x)

Example:
Differentiate f ( x )=5 ( x2 −3)

Solution:
'
f ' ( x )=5 ( x2−3 ) =5 ( 2 x )=10 x
4. Derivatives of Sums and Difference of Functions

If f ( x )=g ( x ) +h(x ), then f ' ( x )=g ' ( x )+ h' ( x). Note that g( x ) and h( x ) are two distinct
differentiable functions.

Example

Given that f ( x )=g ( x ) +h( x ) and g ( x )=x 2−3 and h ( x )=8 x , what is f ' ( x )?

Solution:

Since g' ( x )=2 x and h ' ( x )=8 , then f ' ( x )=2 x +8

5. The Product Rule

If f ( x )=g ( x ) h(x) where g(x) and h(x) are two distinct differentiable functions, then
f ' ( x )=g ( x ) h' x ¿+h ( x ) g ' ( x )

A simple mnemonics can also be used to easily memorize the derivative of the
function that is setting first g(x ) as u and h(x ) as v.

Therefore, in this case, f ' ( x )=udv +vdu

Example:

Solve the derivative of the function f ( x )=( 3 x 2 +2 x−1 ) (2 x+5)

Solution:

Notice that in the equation u=3 x 2 +2 x−1 and v=2 x +5

The formula to be used is f ' ( x )=udv + vdu

 The solution therefore to the problem is f ' ( x )=( 3 x2 +2 x−1 ) ( 2 )+ ( 2 x +5 ) ( 6 x+ 2 )

Simplifying further we have f ' ( x )=6 x 2 +4 x−2+ 12 x 2+ 4 x +30 x+ 10=18 x 2+38 x +8

6. The Quotient Rule

Let g ( x ) and h ( x ) be differentiable functions and h ( x ) is not equal to zero. The

' '
g(x ) ' ( x ) h ( x ) g ( x ) −g ( x ) h (x )
derivative therefore of the function f ( x )= is f = 2 .
h(x) [h ( x )]
vdu−udv
If you let g( x ) as u and h(x ) as v, then f ( x )= 2
v

Example:

x2
Solve the derivative of f ( x )=
3x

Solution:

vdu−udv
We will use the formula f ( x )= , in solving the derivative of the given
v2
example.

2
3 x ( 2 x )−x (3)
The derivative of the function is therefore given by f ' ( x )=
( 3 x )2

'( x ) 6 x 2−3 x 2 3 x 2 3 1
Simplifying further we have f = = 2= =
9 x2 9x 9 3

7. The Derivative of a Reciprocal

1 ' −dv
If y= , where v is a differentiable function of x and not equal to zero, y = 2 .
v v
This rule can be verified using the quotient rule.

Example:

1
Differentiate y= 5
x

Solution:
5
' 5x 5
Applying the formula, the derivative of the function therefore is y = 5 2
= 5
(x ) x
8. The Chain Rule

dy d ' ' dy du
If y=f (u) and u=g ( x), so that y=f [ g ( x ) ], then = ¿ = f ( u ) . g ( x )= =
dx dx du dx
dy dy du
or = .
dx du dx

Example:

dy
Given that y=u 2+u and u=3 x 2−1 , solve .
dx

Solution:

dy dy du
We will be using = . in our solution.
dx du dx

dy du
Since y=u 2+u and u=3 x 2−1 , then =2 u+1 and =6 x .
du dx

dy
=( 2u+ 1 )( 6 x )=( 2 ( 3 x −1 ) +1 ) ( 6 x )=( 6 x −2+ 1 ) ( 6 x )=12 x −6 x
2 2 2
Therefore
dx

9. The General Power Rule

Let y= [u ( x ) } ]n , where u=f (x ) is a differentiable function of x and n is a real

number, then

dy d (n−1) du
= [ u ( x ) } ] =n[u ( x )]
n
or d ( un ) =n un−1 du
dx dx dx

Example:
3
Solve the derivative of the function y=( 5 x 3−2 x )

Solution:

Applying the general power rule, the derivative of the function therefore is
y ' =3 ( 5 x 3−2 x ) ( 15 x 2−2 ) =( 5 x3 −2 x ) (45 x2−6)
LEARNING COMPETENCY:

1. Solving the derivatives of functions using differentiation rules

DIRECTIONS:

This module is designed for independent learning and provides a step-by-step

discussion on the topic Differentiation Techniques. All you need to do is to read, understand
and explore the definitions, illustrations and solutions to the examples presented in this
module. A separate video of the lesson will be uploaded to your Edmodo accounts. Series of
exercises on the topic shall also be given to you in which you are required to answer.

Accomplished worksheets should be forwarded to the teacher through any of the

following: Edmodo, Facebook messenger and gmail.

Should you have questions or clarifications regarding this lesson, please feel free to
contact the author of this learning module, Mr. Michael M. Acupan through messenger Mike
Acupan, gmail michaelmacupan@gmail.com or through his cellphone number 09168388795.

EXERCISES:

Provide neat and complete solutions to the following problems.

Solve the derivative of the functions using differentiation rules.

1. f ( x )=e2 + π
3
2. f ( x )=( 5 x 2−3 ) 4 (2 x)

3. f ( x )=
√
3 3
x

4. f ( x )=( x5 −8 ) (x 2 +5)

( )
5
1
5. f ( x )= 1+
x

GUIDE QUESTIONS

1. What is the reason why the increment method od solving the derivatives of functions is
presented first the different rules?

2. In what way is the general power rule different from the power rule?

3. What operation of function is the basis of the chain rule?

4. What is the importance of knowing the derivatives of functions?

RUBRIC FOR SCORING OF OUTPUTS

For the individual items given in the EXERCISES part of this module, 5 points will be given
to every neat and complete solution. However, if solutions are not complete, partial points
will be credited to you based on the completeness of the solution presented. The teacher will
determine the corresponding partial points to be given.

REFERENCES

The following references might be of help to you in understanding better the concepts
presented in this module

Books:
1. Calculus Made Easy for High School Students by Felipe Commandante
2.The Calculus 7 by Leithold

e-reference/s:

Basic Differentiation Rules- Organic Chemistry

Link
https://www.youtube.com/watch?v=IvLpN1G1Ncg