HIGH SCHOOL
DIFFERENTIAL CALCULUS
COURSE
4.3. DIFFERENTIATION RULES
Differentiation rules allow us to find derivatives without the use of the limit definition. These are found by
identifying certain patterns of differentiation.
Algebraic functions
1. Constant Rule: The derivative of a constant function is always 0.
d
[c ] = 0
dx
Examples:
f ( x) = 3
f ( x) = 0
d
[100] = 0
dx
2. Power Rule: Considering n as a rational number, any function f ( x) = x n :
d n
[ x ] = nx n 1
dx
Examples:
Function
Derivative
3
f ( x) = x
1
g ( x) = 2 = x 2
x
h( x) = x = x1/2
y = x5 = x5/2
f ( x) = 3x31 = 3x 2
d
2
[ g ( x)] = 2 x 3 =  3
dx
x
d
1
1
[h( x)] = x 1/2 =
dx
2
2 x
y =
5 3/2 5 x3
x =
2
2
3. Constant Multiple Rule: When having a constant c (real number) as a coefficient of a function
f ( x ) , doing the derivative of the entire function results in:
d
[cf ( x)] = cf ( x)
dx
the constant can be taken out without change.
Examples:
Function
y = 3x
y=
2
2
= 2 x 1
x
3x
3
= x
2
2
1
1
y=
= x 2/3
3 2
2
2 x
y=
Derivative
dy
d
= 3 x 2 = 3(2 x) = 6 x
dx
dx
dy
d
2
= 2 x 1 = 2(1x 2 ) =  2
dx
dx
x
3 d
3
y = 
x = (1) = 3 / 2
2 dx
2
1 d 2/3
1
y =
(x ) = 
3
2 dx
3 x5
NOTE: Again, rewriting a function is very important when differentiating. Also, use parentheses when
needed.
4. Sum and difference rules: The derivative of a sum or a difference of functions is equal to the sum or
difference of their derivatives.
d
[ f ( x)  g ( x)] = f ( x)  g ( x)
dx
Examples:
f ( x) = x 3  4 x  5
d
d
d
f ( x) = [ x3 ]  [4 x]  [5] = 3 x 2  4  0
dx
dx
dx
x4
 3x3  2 x
2
4
g ( x) =  x 3  3(3) x 2  2 = 2 x 3  9 x 2  2
2
g ( x) = 
5. Product Rule: This rule is not as easy as the sum or difference rule, because it has a special and
more complex form.
The derivative of a product of two functions results in:
d
[ f ( x)  g ( x)] = f ( x)  g ( x)  g ( x)  f ( x)
dx
or, to simplify, considering f ( x) = u and g ( x ) = v :
d
[u  v] = u  v  v  u 
dx
In words, that would be: ``The first function times the derivative of the second function, plus the second
function times the derivative of the first function''
Examples:
Example 1:
f ( x) = (3x  2 x 2 )  (5  4 x)
d
d
f ( x) = (3x  2 x 2 )  [5  4 x]  [3 x  2 x 2 ]  (5  4 x)
dx
dx
2
= (3x  2 x )(4)  (5  4 x)(3  4 x)
= (12  8 x 2 )  (15  20 x  12 x  16 x 2 )
= 24 x 2  4 x  15
Example 2:
g ( x) = (3x 2  3)  (5 x 2  3x)  5 x7
g ( x) = (3x 2  3)  (10 x  3)  (5x 2  3x)  (6 x)  35x 6
= 30 x3  9 x 2  30 x  9  30 x3  18 x 2  35 x 6
= 35 x 6  60 x3  27 x 2  30 x  9
6. Quotient Rule: The derivative of a quotient of two functions results in:
d  f ( x)  g ( x)  f ( x)  f ( x)  g ( x)
=
,
dx  g ( x) 
[ g ( x)]2
g ( x)  0
or, to simplify, considering f ( x) = u and g ( x ) = v :
d  u  v  u   u  v
=
,
dx  v 
v2
v0
In words, that would be: ``The denominator times the derivative of the numerator, minus the numerator
times the derivative of the denominator, all over the square of the denominator.''
Examples:
Example 1:
y=
5x  2
x2  1
d
d
[5 x  2]  (5 x  2) [ x 2  1]
dx
dx
y =
2
2
( x  1)
2
( x  1)(5)  (5 x  2)(2 x)
=
( x 2  1)2
5 x 2  5  10 x 2  4 x
=
( x 2  1)2
5 x 2  4 x  5
= 4
x  2 x2  1
( x 2  1)
Example 2:
1
x
y
x5
3
1

x3 
x
Rewrite : y = 
x( x  5)
3x  1
= 2
x  5x
( x 2  5x)(3)  (3x  1)(2 x  5)
y =
( x 2  5 x) 2
3x 2  15 x  6 x 2  15 x  2 x  5
=
( x 2  5 x) 2
3x 2  2 x  5
=
( x 2  5 x) 2
*Notes:
• Some functions NEED to be rewritten in order to be solved (differentiated) in an easier way.
• Not all fractions need to be solved using the Quotient Rule, because it is best to use the Constant
Multiple Rule or Product Rule. For instance, when the denominator is a constant or a single variable.
Trigonometric functions
The Trigonometric Rules need to be learnt by memorization mostly, since demonstrating them could
prove impractical. However, using a combination of Trigonometric Identities and Quotient Rule, by
knowing the first two rules one can obtain the others.
1. Derivative of sine:
d
[sin x] = cos x
dx
2. Derivative of cosine:
d
[cos x] =  sin x
dx
3. Derivative of tangent:
d
[tan x] = sec 2x
dx
sin x
,
cos x
(cos x)  (cos x)  (sin x)  (  sin x) cos 2x  sin 2x
1
y =
=
=
= sec 2x
2
2
2
(cos x)
x
x
cos
cos
Demonstration: Knowing the trigonometric identity tan x =
4. Derivative of cotangent:
d
[cot x] = csc 2x
dx
cos x
,
sin x
(sin x)  ( sin x)  (cos x)  (cos x) sin 2x  cos 2x
1
y =
=
=
= csc 2x
2
2
2
(sin x)
x
x
sin
sin
Demonstration: Knowing the trigonometric identity cot x =
5. Derivative of secant:
d
[sec x] = sec x tan x
dx
1
,
cos x
(cos x)  (0)  (1)  ( sin x) sin x 1
y =
=

= sec x tan x
(cos x)2
cos x cos x
Demonstration: Knowing the trigonometric identity sec x =
6. Derivative of cosecant:
d
[csc x] =  csc x cot x
dx
1
Demonstration: Knowing the trigonometric identity csc x =
,
sin x
(sin x)  (0)  (1)  (cos x)
cos x 1
y =
=

=  csc x cot x
2
(sin x)
sin x sin x
Examples:
Example 1:
y = 2sin x
y = 2 cos x
Example 2:
f ( x) = x  cos x
f ( x) = 1  (  sin x)
f ' ( x)  1  sin x
Example 3:
g ( x) = x sec x
g ( x) = x(sec x tan x)  (sec x)
Logarithmic functions
The derivative of the natural logarithmic function is the following:
d
1
[ln x] = ,
dx
x
x>0
Sometimes it is best to use the properties of logarithms to rewrite the function before differentiating, for it
will make it easier to find the derivative.
Exponential functions
The derivative of the natural exponential function is the following:
d x
[e ] = e x
dx
Examples:
y = x ln x
1
y = x    ln x = 1  ln x
 x
Sometimes it is better to use the properties of logarithms to rewrite the function before differentiating
(Please refer to the notes about Concepts of Logarithmic functions)
Examples:
Example 1:
f ( x) = ln x  1 = ln( x  1)1/2 =
f ( x) =
1
ln( x  1)
2
1 1
1

=
2 x  1 2( x  1)
Example 2:
x( x  1) 2
x 1
Rewrite  g ( x) = ln[ x( x  1) 2 ]  ln x  1
= ln x  ln( x  1)2  ln( x  1)1 / 2
1
= ln x  2 ln( x  1)  ln( x  1)
2
1
1
1 1
Differentiate  g ( x) =  2 
 
x
x  1 2 x 1
1
2
1
= 

x x  1 2( x  1)
g ( x) = ln
Summarizing, the General procedure to find derivatives would be:
1. Rewrite the function (when having radicals or rational functions)
2. Identify the appropriate Basic Differentiation Rule (1 or more)
3. Find the derivative
4. Simplify when needed
4.3.2 Defining Chain Rule
The Chain Rule is a differentiation rule that is mostly applied to composite functions. It can also be said
that the Chain Rule is a general differentiation rule, which encompasses all previous rules.
Chain Rule:
If y = f (u ) is a differentiable function of u , and u = g ( x ) is a differentiable function of x :
dy dy du
=

dx du dx
or,
d
d
[ f ( g ( x))] = [ f (u )] = f (u )  u 
dx
dx
That is, y will be the derivative of the outside function (with respect to u ) times the derivative of the
inside function.
This can also be extended to composite functions with more than two functions.
Example: If having a function y = h( f ( g ( x))) , taking u = g ( x ) , f (u ) = f ( g ( x)) and v = h( f ( g ( x))) :
y = v  f (u )  u 
... and so on.
Special case 1: General Power Rule
Although we can say that the Chain Rule is universally applied, we could also specify one of the most
common cases, the General Power Rule. This rule follows the basic power rule, as stated in (1.7), but in
this case involves a composite function instead of a simple function on x .
If u = g ( x ) and y = u n ,
d n
[u ] = nu n 1  u 
dx
Examples:
Example 1:
f ( x) = (3x  2 x 2 )3
If u = 3x  2 x 2
and f (u) = u 3 = (3x  2 x 2 )3
f ( x) = 3u 2  u = 3(3x  2 x 2 )2  (3  4 x)
Example 2:
y = ( x3  1)100
If u = x3  1
and f (u) = u100 = ( x3  1)100
y = 100u 99  u = 100( x3  1)99  3x 2 = 300 x 2  ( x3  1)99
Example 3:
g ( x) = (2 x  1)5  ( x3  x  1) 4
g ( x) = (2 x  1)5  4( x3  x  1)3  (3x 2  1)  ( x3  x  1)4  5(2 x  1) 4  2
= 2(2 x  1)4 ( x3  x  1)3 (17 x3  6 x 2  9 x  3)
Example 4:
 t2 
y=

 2t  1 
 t  2  (2t  1)(1)  (t  2)(2)
y = 9 

(2t  1) 2
 2t  1 
9
=
45(t  2)8
(2t  1)10
Special case 2: General Trigonometric Rules:
Chain Rule also applies for Trigonometric functions, but, unlike the rules seen in section 4.3.1, here
instead of having x as an argument, the function has u (that is, g ( x ) ). Therefore, the new Trigonometric
Rules would be:
d
[sin u ] = cos u  u 
dx
d
[cos u ] =  sin u  u 
dx
d
[tan u ] = sec 2u  u 
dx
d
[cot u ] = csc 2u  u 
dx
d
[sec u ] = sec u tan u  u 
dx
d
[csc u ] =  csc u cot u  u 
dx
Examples:
y = sin(cos(tan x))
dy
d
= cos(cos(tan x))  [cos(tan x)]
dx
dx
d
= cos(cos(tan x))  (  sin(tan x))  [tan x]
dx
= cos(cos(tan x))  ( sin(tan x))  (sec 2x)
=  cos(cos(tan x))  sin(tan x)  sec 2x
Special case 3: General natural logarithmic and exponential rules:
Applying the concept of Chain Rule to the existing rules for natural logarithmic and exponential functions
will result in:
d
1
[ln u ] =  u ,
dx
u
and,
d u
[e ] = e u  u 
dx
Examples:
Example 1:
y = ln  2 x 
y =
1
1
2 =
2x
x
Example 2:
f ( x) = x  ln( x 2  1)
 1

f ( x) = x  2
 2 x   ln( x 2  1)
 x 1

=
2 x2
 ln( x 2  1)
2
x 1
Example 3:
g ( x) = e2 x 1
g ( x) = e2 x 1  2 = 2e2 x 1
Example 4:
2
1  x2
h( x ) =
e
2
2
1  x2
2x
h( x) =
e 
2
2
2
x

1
=
xe 2
2
u>0