11.3:Derivatives of Products
and Quotients
Can you find the derivatives of the following functions?
1)
2)
3)
4)
f(x) = x2 x4
f(x) = 3x3 (2x2 – 3x + 1)
f(x) = 5x8 ex
f(x) = x7 ln x
** You can find the derivative
for problem 1 and 2 easily by multiply
two functions, but not for problem
3 and 4.
** You can find the derivative for all
of the above problems using the limit
definition, but it can be a long and
tedious process
As you know in section 10.5,
the derivative of a sum is the
sum of the derivatives.
F(x) = u(x) + v(x)
F’(x) = u’(x) + v’(x)
Is the derivative of a product
the product of the
derivatives?
NO
Derivatives of Products
Theorem 1 (Product Rule)
If f (x) = u(x)  v(x), and if u ’(x) and v ’(x) exist, then
f ’ (x) = u(x)  v ’(x) + u ’(x)  v(x)
In words: 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.
Example 1) Find f’(x) if f(x) = x2 x4
Method 1
f(x) = x2 x4
f(x) = x6
f’(x) = 6x5
Method 2 – Apply product rule
F(x) = x2 x4
F’(x) = x2 (x4)’ + x4 (x2)’
F’(x) = x2 (4x3) + x4 (2x)
F’(x) = 4x5 + 2x5
F’(x) = 6x5
Example 2) Find f’(x) if f(x) = 3x3 (2x2 – 3x + 1)
Method 1
f(x) = 3x3 (2x2 – 3x + 1)
f(x) = 6x5 – 9x4 + 3x3
f’(x)= 30x4 – 36x3 + 9x2
Method 2: Apply product rule
f(x) = 3x3 (2x2 – 3x + 1)
f’(x)= 3x3 (2x2 – 3x + 1)’ + (2x2 – 3x + 1)(3x3)’
f’(x)= 3x3 (4x -3) + (2x2 – 3x + 1)(9x2)
f’(x)= 12x4 – 9x3 +18x4 – 27x3 + 9x2
f’(x)= 30x4 – 36x3 + 9x2
Example 3) Find f’(x) if f(x) = 5x8 ex
f(x) = 5x8 ex
f’(x) = 5x8 (ex)’ + ex (5x8)’
f’(x) = 5x8 (ex) + ex (40x7)
f’(x) = 5x8 ex + 40 x7 ex
f’(x) = 5x7 ex (x + 8) or 5x7 (x + 8) ex
Note that the only way to do is to apply the product rule
Example 4) Find f’(x) if f(x) = x7 ln x
f(x) = x7 ln x
f’(x) = x7 (ln x)’ + ln x (x7)’
f’(x) = x7 (1/x) + ln x (7x6)
f’(x) = x6
+ 7x6 ln x
f’(x) = x6 (1+ 7 lnx)
Example 5
Let f(x) = (2x+9)(x2 -12)
A) Find the equation of the line tangent to the graph of f(x) at x = 3
f’(x) = (2x+9)(2x) + (x2 – 12)(2)
f’(x) = 4x2 + 18x + 2x2 – 24 = 6x2 + 18x - 24
so slope m = f’(3) = 6(3)2 + 18(3) – 24 = 84
also, if x = 3, f(3) = (2*3+9)(32 -12) = -45
y = mx + b
-45 = 84(3) + b
b = - 297
Therefore the equation of the line tangent is y = 84x - 297
B) Find the values(s) of x where the tangent line is horizontal
The slope of a horizontal line is 0 so set f’(x) = 0
6x2 + 18x – 24 = 0
6(x2 + 3x – 4) = 0
6(x + 4) (x – 1) = 0 so x = -4 or 1
Derivatives of Quotients
Theorem 2 (Quotient Rule)
If f (x) = T (x) / B(x), and if T ’(x) and B ’(x) exist, then
f '( x) 
T '( x)  B ( x)  B '( x) T ( x) 
[ B ( x )]
2
In words: 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.
Example 6
A) Find f’(x) for
f ( x) 
2x
x 3
2
( 2 x )' ( x  3 )  ( x  3 )' ( 2 x )
2
f '( x) 
2
x
2
 3
2
2 ( x  3 )  ( 2 x )( 2 x )
2
f '( x) 
x
2
 3
2
2x  6  4x
2
f '( x) 
x
2
 3
2
2
f '(x) 
6  2x
x
2
2
 3
2
Example 6 (continue)
t  3t
3
B) Find f’(x) for f ( x ) 
t 4
2
( t  3 t )' ( t  4 )  ( t  4 )' ( t  3 t )
3
f '( x) 
2
(t  4 )
2
2
2
3
(t  4 )
2
2
3 t  12 t  3 t  12  2 t  6 t )
4
f '(x) 
3
( 3 t  3 )( t  4 )  ( 2 t )( t  3 t )
2
f '( x) 
2
2
(t  4 )
2
t  9 t  12
4
f '(x) 
2
2
(t  4 )
2
2
4
2
2
Example 6 (continue)
C) Find d/dx for f ( x ) 
2 x
x
Method 1:
f ( x) 
3
2

3
f ( x)  2 x
3
Method 2: apply quotient rule
2 x
x
3
x
3
3
1
1
f ( x) 
f '(x) 
2 x
3
3
x
2
3
2
3
3 x ( x )  ( 3 x )( 2  x )
3
(x )
3x  6 x  3x
5
f ' ( x)  6 x
f '( x) 
4
6
x
4
f '( x) 
2
f '( x) 
2
 6x
x
6
x
6
2

6
x
4
5
Example 7
A) Find f’(x) for
f ( x) 
x
3
e 2
x
( x )' ( e  2 )  ( e  2 )' x
3
f '( x) 
x
(e  2)
x
x
x
(e  2)
x
2
3x e  6x  e x
2
f '( x) 
2
3 x (e  2)  (e ) x
2
f '( x) 
x
x
2
(e  2)
x
Note that the only way to do is to apply the quotient rule
x
2
3
3
3
Example 7 (continue)
B) Find f’(x) for
4x
f ( x) 
f '( x) 
1  ln x
( 4 x )' (1  ln x )  (1  ln x )' ( 4 x )
f '( x) 
1  ln x 2
4 (1  ln x )  (1 / x )( 4 x )
f '( x) 
f '( x) 
1  ln x 2
4  4 ln x  4
1  ln x 2
4 ln x
1  ln x 2
Example 8
The total sales S (in thousands of games) t months after the game
is introduced is given by S ( t ) 
150 t
t3
A) Find S’(t)  150 ( t  3 )  (1)(150 t )  150 t  450  150 t 
2
2
(t  3)
(t  3)
450
(t  3)
2
B) Find S(12) and S’(12). Explain
S(12) = 120
and S’(12) = 2
After 12 months, the total sales are 120,000 games. Sales are
increasing at a rate of 2000 games per month.
C) Use the results above to estimate the total sales after 13 months
S(13) = S(12) + S’(12) = 122. The total sales after 13 months is
122,000 games.