2.3 The Product and Quotient Rules
d
 f  x  g  x   f  x  g '  x   g  x  f '  x 
dx 
Chant: the first times the derivative of the second plus the second times the derivative of the first
The Product Rule:
1. Find the derivative: h  x    3x  2 x 2   5  4 x 
[FOIL first and do it the old way, then use the product rule]
2. Find the derivative: y  x sin x
3. Find the derivative: y  2x cos x  2sin x
[treat each term as a separate problem]
1
d  f  x  g  x f ' x  f  x g ' x


2
dx  g  x  
 g  x  
Chant: the bottom times the derivative of the top minus the top times the derivative of the bottom
all over bottom squared (or LO di HI minus HI di LO all over LO squared)
The Quotient Rule:
4. Find the derivative: y 
5x  2
x2  1
5. Find the derivative: y 
3  1/ x 
x5
6. Distinguishing between the quotient rule and the constant multiple rule. Find each derivative.
x 2  3x
a. y 
6
b. y 
c. y 
d. y 
5x4
8
3  3x  2 x 2 
7x
9
5x2
HW: p. 124 #s 1-17odd (check with BOB)
2
2.3 Derivatives of Trigonometric Functions
Recall, from 2.2 that
Here are the other 4:
d
sin x  cos x
dx
d
 tan x  sec2 x
dx
d
sec x  sec x tan x
dx
and
and
and
7. Prove
d
 tan x  sec2 x using the quotient rule.
dx
8. Prove
d
sec x  sec x tan x using the quotient rule.
dx
9. Find each derivative:
a. y  x  tan x
b. y  x sec x
3
d
cos x   sin x
dx
d
cot x   csc2 x
dx
d
csc x   csc x cot x
dx
10. Show that the derivatives of both expressions are equal. (just take the derivative of both sides)
1  cos x
y
 csc x  cot x
sin x
Often, you will need to simplify (sometimes quite a bit) after differentiation to get the final answer. See chart
on the bottom of p. 122.
For certain problems you will need to differentiate more than once. These are called high order derivatives.
They are notated using y '' (y double prime), f ''  x  , y ''', y 4 , etc…
x2  2 x 1
11. Find the second derivative of y 
(#86 from text)
x
12. Find the second derivative of y  sec x (#88 from text)
13. Given f  4  x   2 x  1 , find f  6  x  (#92 from text)
HW: p. 124 #s 39-45odd, 49, 51; p. 125 #s 83-91odd (check with BOB)
4
2.3 Continued: Acceleration and other nifty stuff
In section 2.2, you learned that the derivative of the position function was the velocity function. Well, the
second derivative of the position function is the acceleration function. Acceleration is the rate of change of
the velocity.
s  t  is the position function
v  t   s '  t  is the velocity function
a  t   v '  t   s ''  t  is the acceleration function
14. The velocity of an object in meters per second is: v  t   36  t 2 , 0  t  6 . Find the velocity and
acceleration when t  3 . What can be said about the speed of the object when the velocity and
acceleration have opposite signs? (#101 from text)
15. In point-slope form, find the equation of the tangent line to the graph of f at the indicated point.
x 1
 1
a. f  x  
;
 2,  (#66 from text)
x 1
 3
b. f  x   sec x;
π 
 , 2  (#68 from text)
3 
16. Determine the point(s) at which the graph of y 
x2
has a horizontal tangent. (#70 from text)
x2  1
HW: pp. 124-126 #s 35, 63, 65, 67, 69, 102, 103
5
Product and Quotient Rule Activity
If f and g are the functions whose graphs are shown, let u  x   f  x  g  x  and v  x   f  x  / g  x  .
(a) Find u ' 1 .
(b) Find v '  5 .
y
g


f
x




Let P  x   F  x  G  x  and Q  x   F  x  / G  x  , where F and G are the functions whose graphs are shown.
(a) Find P '  2  .
(b) Find Q '  7  .
y

F

Gx




6