Basic Rules of Differentiation
By
Dr. Julia Arnold and Ms. Karen
Overman
using Tan’s 5th edition Applied
Calculus for the managerial , life,
and social sciences
Basic Rules of Differentiation
Using the definition of the
f x  h   f x 
derivative,
f '  x   lim
h 0
h
can be tiresome.
In this lesson we are going to learn
and use
some basic rules of differentiation
that are
derived from the definition. For
these rules,
let’s assume that we are discussing
differentiable functions.
Rule 1: Derivative of a constant
d c 
0
dx
, for any constant,c.
This rule states that the derivative of a constant is zero.
For example,
f ( x)  5
f ' ( x)  0
Rule 2: The Power Rule
 
d xn
 nxn 1
dx
, where n is any real number
This rule states that the derivative of x raised to a power is the
power times x raised to a power one less or n-1.
For example,
f x   x 5
f ' x   5 x 4
Notice that the derivative is the original power, 5 times x
raised to the fourth, which is one less than 5.
Rule 3: Derivative of a Constant Multiple of a Function
d
cf ( x)  c d  f ( x) , where c is a constant
dx
dx
This rule states that the derivative of a constant times a
function is the constant times the derivative of the function.
For example, find the derivative of
 
d
 f x   d 5 x 4
dx
dx
d 4
5
x
dx
 5 4 x3
 
 
 20x 3
f x   5x 4
Rule 4: Derivative of a Sum or Difference
d
d
d
 f ( x)  g ( x)   f ( x)  g ( x)
dx
dx
dx
This rule states that the derivative of a sum or difference is
the sum or difference of the derivatives.
For example, find the derivative of x2 +2x -3


 
d 2
d 2
d
d
x  2x  3 
x  2 x   3
dx
dx
dx
dx
 2x  2  0
 2x  2
The derivative of x squared is done by the Power Rule (2), the
derivative of 2x is done by rule 3 and power rule and the derivative
of 3, a constant is 0.
More Examples: Find
d
dx
 x
You’ll notice none of the basic rules specifically mention
radicals, so you should convert the radical to its exponential
form, X1/2 and then use the power rule.
d
dx
 
1
1
1

1


d 2
1 2
1 2
1
1


x  x   x  x  1 
dx   2
2
2 x
2
2x
More Examples: Find
d  1 
 2
dx  x 
Again, you need to rewrite the expression so that you can use
one of the basic rules for differentiation. If we rewrite the
fraction as x-2 ,then we can use the power rule.
d  1  d 2
2
 21
3
x  2 x  2 x  3
 2
dx  x  dx
x
 
More Examples: Find
d  4 x3  2 x  7 


dx 
x

Rewrite the expression so that you can use the basic rules of
differentiation.
4 x3  2 x  7 4 x3 2 x 7


  4 x 2  2  7 x 1
x
x
x
x
Now differentiate using the basic rules.
d  4 x3  2 x  7  d

 
4 x 2  2  7 x 1
dx 
x
 dx





d
d
2   d 7 x 1
4x2 
dx
dx
dx
 4  2 x  0  7  1x 11

 8 x  7 x 2
 8x 
7
x2

Another example: Find the slope and equation of the tangent
line to the curve y  2 x 2  1 at the point (1,3).
Recall from the previous chapter, the derivative gives the slope
of the tangent to the curve. So we will need to find the
derivative and evaluate it at x = 1 to find the slope
at the point (1,3). Then we’ll use the slope and the
point to write the equation of the tangent line
using the point slope form.
Find the slope and equation of the tangent line to the curve
y  2 x 2  1 at the point (1,3).
First find the derivative of the function.


 
d
 y   d 2 x 2  1  d 2 x 2  d 1
dx
dx
dx
dx
 2  2 x 21  0
 4x
Now, evaluate the derivative at x = 1 to find the slope at (1,3).
m  4 1  4
Example continued: Now we have the slope of the tangent line at
the point (1,3) and we can write the equation of the line.
Recall to write the equation of a line, start with the point slope form
and use the slope, 4 and the point (1,3).
y  y1  m x  x1 
y  3  4 x  1
y  3  4x  4
y  4x 1
The graph below shows the curve y  2 x 2  1 in blue and the tangent
line at the point (1,3),
y  4 x  1 in red.
Another example: Find all the x values where y  x3  2x 2  x
has a horizontal tangent line.
Find the derivative.


d 3
x  2 x 2  x  3x 2  4 x  1
dx
Since horizontal lines have a slope of 0, set the derivative equal
to 0 and solve for x.
Thus the x values where
the function has horizontal
3x 2  4 x  1  0
tangents is at x = -1 , -1/3.
3x  1x  1  0
3x  1  0 or x  1  0
1
x
or x  1
3
The graph below shows the function from the last example and the
horizontal tangent lines at x=-1 and x=-1/3.