1. 11.3:Derivatives of Products and Quotients

2. NO Can you find the derivatives of the following functions?
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

3. 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.

4. 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) = 4x x5
F’(x) = 6x5

5. 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

6. Example 3) Find f’(x) if f(x) = 5x8 ex
f’(x) = 5x8 (ex)’ + ex (5x8)’
f’(x) = 5x8 (ex) + ex (40x7)
f’(x) = 5x8 ex 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

7. Example 4) Find f’(x) if f(x) = x7 ln x
f’(x) = x7 (ln x)’ + ln x (x7)’
f’(x) = x7 (1/x) + ln x (7x6)
f’(x) = x x6 ln x
f’(x) = x6 (1+ 7 lnx)

8. 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
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

9. Derivatives of Quotients
Theorem 2 (Quotient Rule)
If f (x) = T (x) / B(x), and if T ’(x) and B ’(x) exist, then
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.

10. Example 6
A) Find f’(x) for

11. Example 6 (continue)
B) Find f’(x) for

12. Example 6 (continue) C) Find d/dx for Method 1:
Method 2: apply quotient rule

13. Example 7 A) Find f’(x) for
Note that the only way to do is to apply the quotient rule

14. Example 7 (continue)
B) Find f’(x) for

15. Example 8
The total sales S (in thousands of games) t months after the game
is introduced is given by
A) Find S’(t)
B) Find S(12) and S’(12). Explain
S(12) = 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.