§3.2 Some Differentiation Formulas
The student will learn about derivatives
of constants, powers, sums and differences,
notation, the product rule,
the quotient rule, and
the chain rule
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The Derivative of a Constant
Let y = f (x) = C be a constant function, then
y’ = f ’ (x) = 0.
What is the slope of a constant function?
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Example 1
f (x) = 17
If y = f (x) = C then y’ = f ’ (x) = 0.
f ‘ (x) = 0
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Power Rule.
A function of the form f (x) = xn is called a
power function. (Remember √x and all radical
functions are power functions.)
Let y = f (x) = xn be a power function, then
y’ = f ’ (x) = n xn – 1.
THIS IS VERY IMPORTANT. IT WILL BE
USED A LOT!
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Example 2
f (x) = x5
If y = f (x) = xn then y’ = f ’ (x) = n xn – 1.
f ‘ (x) = 5 • x4 = 5 x4
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Example 3
f (x) =
3
x
f (x) = 3 x , can be rewritten as f (x) = x1/3 and
we can then find the derivative.
f (x) = x 1/3
f ‘ (x) = 1/3 x - 2/3
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Constant Multiple Property.
Let y = f (x) = k • u (x) be a constant k times a
differential function u (x). Then
y’ = f ’ (x) = k • u’ (x) = k • u’.
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Example 4
f (x) = 7x4
If y = f (x) = k • u (x) then f ’ (x) = k • u’.
f ‘ (x) = 7 • 4 • x3 = 28 x3
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Emphasis
f (x) = 7x
If y = f (x) = k • u (x) then f ’ (x) = k • u’.
f ‘ (x) = 7 • 1 = 7
REMINDER: If f ( x ) = c x then f ‘ ( x ) = c
The derivative of x is 1.
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Sum and Difference Properties.
• 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.
OR
If y = f (x) = u (x) ± v (x), then
y ’ = f ’ (x) = u ’ (x) ± v ’ (x).
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Example 5
f (x) = 3x5 + x4 – 2x3 + 5x2 – 7 x + 4
From the previous examples we get f ‘ (x) = 15x4 + 4x3 – 6x2 + 10x – 7
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Example 6
f (x) = 3x - 5 - x - 1 + x 5/7 + 5x- 3/5
f ‘ (x) = - 15x - 6 + x - 2 + 5/7 x – 2/7 - 3 x – 8/5
Show how to do fractions on a calculator.
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Notation
Given a function y = f ( x ), the following
are all notations for the derivative.
y′
d
f ( x)
dx
f′(x)
dy
dx
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Graphing Calculators
Most graphing calculators have a built-in
numerical differentiation routine that will
approximate numerically the values of f ’ (x)
for any given value of x.
Some graphing calculators have a built-in
symbolic differentiation routine that will
find an algebraic formula for the derivative,
and then evaluate this formula at indicated
values of x.
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Example 7
f (x) = x 2 – 3x and f ’ (x) = 2x - 3
3. Do the above using a graphing calculator.
Let x = 2.
Using dy/dx under the
Using tangent under
“calc” menu.
the “draw” menu.
slope
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Tangent equation
Example 8 - TI-89 ONLY
f (x) = 2x – 3x2 and f ’ (x) = 2 – 6x
Do the above using a graphing calculator
with a symbolic differentiation routine.
Using algebraic differentiation under the home
“calc” menu.
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Median Summary.
If f (x) = C then f ’ (x) = 0.
If f (x) = xn then f ’ (x) = n xn – 1.
If f (x) = k • u (x) then f ’ (x) = k • u’ (x)
= k • u’.
If f (x) = u (x) ± v (x), then
f ’ (x) = u’ (x) ± v’ (x).
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Derivates of Products
The derivative of the product of two functions is
the first function times the derivative of the
second function plus the second function times
the derivative of the first function.
Product Rule
d
[ f ( x)  s ( x) ]  f ( x)  s ' ( x)  s ( x)  f ' ( x)
dx
OR
d
( f  s )  f  s'  s  f '
dx
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Example
Find the derivative of y = 5x2(x3 + 2).
Product Rule
d
[ f ( x)  s ( x) ]  f (x)  s ' (x)  s ( x)  f ' (x)]
dx
Let f (x) = 5x2 then f ‘ (x) = 10x
Let s (x) = x3 + 2 then s ‘ (x) = 3x2, and
y ‘ (x) = 5x2 • 3x2 + (x3 +
+ 2)
2) •10x
= 15x4 + 10x4 + 20x = 25x4 + 20x
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Derivatives of Quotients
The derivative of the quotient of two functions is
the bottom function times the derivative of the
top function minus the top function times the
derivative of the bottom function, all over the
bottom function squared.
Quotient Rule:
d  t ( x)  b ( x)  t ' ( x)  t ( x)  b ' ( x)

 
dx  b ( x) 
[ b ( x) ] 2
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Derivatives of Quotients
May also be expressed as -
d
dx
 t ( x)  b t'  t b'

 
2
b
 b ( x) 
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Example
Find the derivative of
3x
y
.
2x  5
 t ( x)  b t'  t b'

 
2
b
 b ( x) 
Let t (x) = 3x and then t ‘ (x) =
3.
Let b (x) = 2x + 5 and then b ‘ (x) =
d
dx
2.
15
( 2x  5)  3  3x  2
f ' ( x) 

2
2
( 2x  5)
( 2x  5)
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Median Summary.
Product Rule. If f (x) and s (x), then
d
f  s   f • s ' + s • f '
dx
Quotient Rule. If t (x) and b (x), then
d
dx
t

b
 b t'  t b'
 
2
b

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Composite Functions
Definition. A function m is a composite of
functions f and g if
m (x) = f [ g (x)]
The domain of m is the set of all numbers x such
that x is in the domain of g and g (x) is in the
domain of f.
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Examples
Let f (u) = u4, g (x) = 2x + 5, and m (v) = ln v. Find:
f [ g (x)] = f (2x + 5) = (2x + 5)4
g [ f (x)] = g (x4) = 2x 4 + 5
m [ g (x)] = m (2x + 5) = ln (2x + 5)
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Chain Rule: Power Rule.
We have already made extensive use of the power rule with xn,
d n
n 1
x  nx
dx
We wish to generalize this rule to cover [u (x)]n.
That is, we already know how to find the derivative of
f (x) = x 5
We now want to find the derivative of
f (x) = (3x 2 + 2x + 1) 5
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Chain Rule: Power Rule.
General Power Rule. [Chain Rule]
Theorem 1. If u (x) is a differential function, n
is any real number, and
If f (x) = [u (x)]n
I use u (x) because !!!
then
f ’ (x) = n un – 1 u’
or
d n
n  1 du
u  nu
dx
dx
* * * * * VERY IMPORTANT * * * * *
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Example 1
Find the derivative of y = (x3 + 2) 5.
NOTE: If we let u = x 3 + 2, then y = u 5.
Chain Rule
d n
n  1 du
u  nu
dx
dx
Let u (x) = x3 + 2, then y = u 5 and du/dx = 3x2
d
3
5
( x  2)  5 (x3 + 2)4 3x2
dx
= 15x2(x3 + 2)4
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Examples
Find the derivative of:
y = (x + 3) 2
y’ = 2 (x + 3) (1) = 2 (x + 3)
y = (4 – 2x 5) 7
y’ = 7 (4 – 2x 5) 6 (- 10x 4)
y’ = - 70x 4 (4 – 2x 5) 6
y = 2 (x3 + 3) – 4
y’ = - 8 (x3 + 3) – 5 (3x 2)
y’ = - 24x 2 (x3 + 3) – 5
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Example 2
Find the derivative of y = x 3  3
Rewrite as y = (x 3 + 3) 1/2
Then y’ = 1/2 (x 3 + 3) – 1/2 (3x2)
3 2 3
= x (x + 3) –1/2
2
Try y = (3x 2 - 7) - 3/2
y’ = (- 3/2) (3x 2 - 7) - 5/2 (6x)
= (- 9x) (3x 2 - 7) - 5/2
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Example 3
Find f ’ (x) if f (x) =
x4
( 3x  8)
2
.
We will use a combination of the quotient rule and
the chain rule.
Let the top be t (x) = x4, then t ‘ (x) = 4x3
Let the bottom be b (x) = (3x – 8)2, then using the
chain rule b ‘ (x) = 2 (3x – 8) 3 = 6 (3x – 8)
f ' ( x) 
( 3x  8) 2 ( 4x 3 )  x 4 6 ( 3x  8)
(( 3x  8)2 )2
(3x  8)(4x 3 )  6x4
f '(x) 
3
(3x  8)
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Summary.
Product Rule. If f (x) and s (x), then
d
f  s   f • s ' + s • f '
dx
Quotient Rule. If t (x) and b (x), then
d
dx
t

b
 b t'  t b'
 
2
b

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Summary.
If
y = f (x) = [u (x)]n
then
d n
n  1 du
u  nu
dx
dx
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ASSIGNMENT
§3.2: Page 52; 1 – 23 odd.
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