10.5 Basic Differentiation
Properties
10.5 Basic Differentiation
Properties
• Instead of finding the limit of the different
quotient to obtain the derivative of a
function, we can use the rules of
differentiation (shortcuts to find the
derivatives).
Review: Derivative Notation
There are several widely used symbols to represent
the derivative. Given y = f (x), the derivative may be
represented by any of the following:
■ f ’(x)
■ y’
■ dy/dx
What is the slope of a constant function?
The graph of f (x) = C is a
horizontal line with slope 0, so
we would expect f ’(x) = 0.
Theorem 1. Let y = f (x) = C be a constant function, then
y’ = f ’(x) = 0.
Examples: Find the derivatives of the
following functions
1) f(x) = -24
f’(x) = ? 0
2) f(x) =  7
f’(x) = ? 0
3) f(x) = π
f’(x) = ?
0
Power Rule
A function of the form f (x) = xn is called a power function.
This includes f (x) = x (where n = 1) and radical functions
(fractional n).
Theorem 2. (Power Rule) Let y = xn be a power function,
then
y’ = f ’(x) = n xn – 1.
Example
Differentiate f (x) = x10.
Solution:
By the power rule, the derivative of xn is n xn–1.
In our case n = 10, so we get f ’(x) = 10 x10-1
= 10 x9
Example
Differentiate
f ( x)  3 x.
Solution:
Rewrite f (x) as a power function, and apply the power rule:
f ( x)  x
1/ 3
1 2 / 3
1
f ' ( x)  x 
2
3
3
3 x
Examples: Find the derivatives of the
following functions
1)
x6
f(x) =
f’(x) = ?
4) f(x) =
6x5
2) f(x) = t-2
f’(x) = ? -2t-3=
3) f(x) =
f’(x) =
2
t3
5) f(x) =
?
3
u2
3
t
2
1
2
 1x
2
1
 2
x
= u2/3
2 3
2
u  1/ 3
3
3u
6) f(x) = 1
x
f’(x) =?
= x-1
1
f’(x) =?
t3/2
f’(x) = ?
1
x
= x-1/2
3
1
1
 x 2   3/ 2
2
2x
Constant Multiple Property
Theorem 3. Let y = f (x) = k u(x) be a constant k times a
function u(x). Then
y’ = f ’(x) = k  u’(x).
In words: The derivative of a constant times a function is the
constant times the derivative of the function.
Example
Differentiate f (x) = 10x3.
Solution:
Apply the constant multiple property and the power rule.
f ’(x) = 10(3x2) = 30 x2
Examples: Find the derivatives of the
following functions
1) f(x) = 4x5
f’(x) = ?
20x4
4) f(x) =
t4
12
f’(x) = 1 t 3
3
1 3
2) f(x) = 1
x
3 
3x
3
f’(x) = ?
1
4
 1x  4
x
1

0 .9
0.9
5) f(x) = 3
 1  0.9 x 3
x
3
x
f’(x) =
 0.3x

4
3
3

10x 4 / 3
Sum and Difference Properties
Theorem 5. If
y = f (x) = u(x) ± v(x),
then
y’ = f ’(x) = u’(x) ± v’(x).
In words:
■ The derivative of the sum of two differentiable functions is
the sum of the derivatives.
■ The derivative of the difference of two differentiable
functions is the difference of the derivatives.
Example
Differentiate f (x) = 3x5 + x4 – 2x3 + 5x2 – 7x + 4.
Solution:
Apply the sum and difference rules, as well as the constant
multiple property and the power rule.
f ’(x) = 15x4 + 4x3 – 6x2 + 10x – 7.
Examples: Find the derivatives of the
following functions
1)
3x4-2x3+x2-5x+7
f(x) =
f’(x) = ?
12x  6 x  2 x  5
3
2
2) f(x) = 3 - 7x -2
f’(x) = ?
14
14 x  3
x
3
4) f(x) = 5v
f’(x) =?
3
 v  5v  v
4
3
1 3 / 4
1
2
15v  v
 15v  3 / 4
4
4v
2
3
4 x4
 3
5) f(x) = 
4x x
8
4
3 1
x
  x  4 x 3 
4
8
3 2
1 3
4
f’(x) =? 4 x  12 x  2 x
3 12 1 3
 4 x
2
4x
x
2
1/ 4
Applications
Remember that the derivative gives the instantaneous rate of
change of the function with respect to x. That might be:
■ Instantaneous velocity.
■ Tangent line slope at a point on the curve of the function.
■ Marginal Cost. If C(x) is the cost function, that is, the total cost
of producing x items, then C’(x) approximates the cost of
producing one more item at a production level of x items. C’(x) is
called the marginal cost.
Example: application
Let f (x) = x4 - 8x3 + 7
(a) Find f ’(x)
(b) Find the equation of the tangent line at x = 1
(c) Find the values of x where the tangent line is horizontal
Solution:
a)
f ’(x) = 4x3 - 24x2
b)
Slope: f ’(1) = 4(1) - 24(1) = -20
Point: (1, 0)
y = mx + b
0 = -20(1) + b
20 = b
So the equation is y = -20x + 20
c) Since a horizontal line has the slope of 0, we must solve f’(x)=0 for x
4x3 - 24x2 = 0
4x2 (x – 6) = 0
So x = 0 or x = 6
Example: more application
The total cost (in dollars) of producing x radios per day is
C(x) = 1000 + 100x – 0.5x2
for 0 ≤ x ≤ 100.
1. Find the marginal cost at a production level of radios.
Solution: C’(x) = 100 – x
2) Find the marginal cost at a production level of 80 radios and
interpret the result.
Solution: The marginal cost is
C’(80) = 100 – 80 = $20
It costs $20 to produce the next radio (the 81st radio)
3) Find the actual cost of producing the 81st radio and compare
this with the marginal cost.
Solution: The actual cost of the 81st radio will be
C(81) – C(80) = $5819.50 – $5800 = $19.50.
Summary
• If f (x) = C, then f ’(x) = 0
• If f (x) = xn, then f ’(x) = n xn-1
• If f (x) = k xn then f ’ (x) = kn xn-1
• If f (x) = u(x) ± v(x),
then f ’(x) = u’(x) ± v’(x).