Clear simple and introduction to further levels, stimulating our passion on calculus basics differential equations being part of them.

1. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 1 of 33
Chapter 3
Techniques of Differentiation

2. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 2 of 33
 The Product and Quotient Rules
 The Chain Rule and the General Power Rule
 Implicit Differentiation and Related Rates
Chapter Outline

3. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 3 of 33
§ 3.1
The Product and Quotient Rules

4. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 4 of 33
 The Product Rule
 The Quotient Rule
 Rate of Change
Section Outline

5. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 5 of 33
The Product Rule

6. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 6 of 33
The Product Rule
EXAMPLE
SOLUTION
Differentiate the function.
Let and . Then, using the product rule, and the
general power rule to compute g΄(x),
  10
2
2
3
3 
 x
x
  3
2

 x
x
f    10
2
3

 x
x
g
  
     
     
3
3
3
3
3
3 2
10
2
10
2
2
10
2
2









 x
dx
d
x
x
dx
d
x
x
x
dx
d
        x
x
x
dx
d
x
x 2
3
3
3
10
3
10
2
2
9
2
2









      .
2
3
2
3
10
3
10
2
9
2
2
x
x
x
x
x 








7. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 7 of 33
The Quotient Rule

8. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 8 of 33
The Quotient Rule
EXAMPLE
SOLUTION
Differentiate.
Let and . Then, using the quotient rule
x
x
x 3
4 2
4


  3
4 2
4


 x
x
x
f   x
x
g 
     
2
2
4
2
4
2
4 3
4
3
4
3
4
x
x
dx
d
x
x
x
x
dx
d
x
x
x
x
dx
d













 

   
2
2
4
3
1
3
4
8
4
x
x
x
x
x
x 






2
2
2
2
2
2
4
2
2
4
3
4
3
3
4
3
3
4
3 








 x
x
x
x
x
x
x
x
x
x
Now simplify.

9. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 9 of 33
The Quotient Rule
Now let’s differentiate again, but first simplify the expression.
x
x
x
x
x
x
x
x 3
4
3
4 2
4
2
4





Now we can differentiate the function in its new form.
CONTINUED
1
3
3
4 


 x
x
x
  2
2
1
3
3
4
3
3
4 





 x
x
x
x
x
dx
d
Notice that the same answer was acquired both ways.

10. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 10 of 33
Rate of Change
EXAMPLE
SOLUTION
(Rate of Change) The width of a rectangle is increasing at a rate of 3 inches
per second and its length is increasing at the rate of 4 inches per second. At
what rate is the area of the rectangle increasing when its width is 5 inches and
its length is 6 inches? [Hint: Let W(t) and L(t) be the widths and lengths,
respectively, at time t.]
Since we are looking for the rate at which the area of the rectangle is changing,
we will need to evaluate the derivative of an area function, A(x) for those given
values (and to simplify, let’s say that this is happening at time t = t0). Thus
     
t
W
t
L
t
A 
 This is the area function.
         
t
L
t
W
t
W
t
L
t
A 





 Differentiate using the product
rule.

11. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 11 of 33
Rate of Change
Now, since the width of the rectangle is increasing at a rate of 3 inches per
second, we know W΄(t) = 3. And since the length is increasing at a rate of 4
inches per second, we know L΄(t) = 4.
Now we substitute into the derivative of A.
This is the derivative function.
         
t
L
t
W
t
W
t
L
t
A 






CONTINUED
W΄(t) = 3, L΄(t) = 4, W(t) = 5,
and L(t) = 6.
       
4
5
3
6
0 

 t
A
Simplify.
  sec
/
in
38 2
0 
 t
A
Now, we are determining the rate at which the area of the rectangle is
increasing when its width is 5 inches (W(t) = 5) and its length is 6 inches
(L(t) = 6).

12. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 12 of 33
The Product Rule & Quotient Rule
Another way to order terms in the product and quotient rules, for the purpose of
memorizing them more easily, is
   
         
x
f
x
g
x
g
x
f
x
g
x
f
dx
d




PRODUCT RULE
 
 
       
 
 
.
2
x
g
x
f
x
g
x
g
x
f
x
g
x
f
dx
d 









QUOTIENT RULE

13. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 13 of 33
§ 3.2
The Chain Rule and the General Power Rule

14. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 14 of 33
 The Chain Rule
 Marginal Cost and Time Rate of Change
Section Outline

15. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 15 of 33
The Chain Rule

16. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 16 of 33
The Chain Rule
EXAMPLE
SOLUTION
Use the chain rule to compute the derivative of f(g(x)), where
and .
Finally, by the chain rule,
  2
4
x
x
x
f 

  4
1 x
x
g 

  ,
2
4
2
x
x
x
f 


   3
4x
x
g 


 
 
 
 
4
2
4
1
2
1
4
x
x
x
g
f 





 
   
   
 
   .
4
1
2
1
4 3
4
2
4
x
x
x
x
g
x
g
f
x
g
f
dx
d


















17. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 17 of 33
The Chain Rule
EXAMPLE
SOLUTION
Compute using the chain rule.
Since y is not given directly as a function of x, we cannot compute by
differentiating y directly with respect to x. We can, however, differentiate with
respect to u the relation , and get
dx
dy
2
2
,
1 x
u
u
y 


dx
dy
1

 u
y
.
1
2
1


u
du
dy
Similarly, we can differentiate with respect to x the relation and get
2
2x
u 
.
4x
dx
du


18. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 18 of 33
The Chain Rule
Applying the chain rule, we obtain
.
4
1
2
1
x
u
dx
du
du
dy
dx
dy




It is usually desirable to express as a function of x alone, so we substitute
2x2 for u to obtain
CONTINUED
dx
dy
.
1
2
2
4
2


x
x
dx
dy

19. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 19 of 33
Marginal Cost & Time Rate of Change
EXAMPLE
SOLUTION
(Marginal Cost and Time Rate of Change) The cost of manufacturing x cases of
cereal is C dollars, where . Weekly production at t weeks from
the present is estimated to be x = 6200 + 100t cases.
.
dx
dC
2
4
3 

 x
x
C
(a) Find the marginal cost,
(b) Find the time rate of change of cost,
(c) How fast (with respect to time) are costs rising when t = 2?
.
dt
dC
  x
x
x
dx
d
dx
dC 2
3
2
4
3 




(a) We differentiate C(x).

20. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 20 of 33
Marginal Cost & Time Rate of Change
(b) To determine , we use the Chain Rule.
dt
dC
  100
100
6200
,
2
3 



 t
dt
d
dt
dx
x
dx
dC
Now we rewrite x in terms of t using x = 6200 + 100t.
CONTINUED
 
100
2
3 








x
dt
dx
dx
dC
dt
dC
 
100
100
6200
2
3 








t
dt
dC
(c) With respect to time, when t = 2, costs are rising at a rate of
 
  .
per week
cereal
of
cases
5
.
302
100
2
100
6200
2
3
2













t
dt
dC

21. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 21 of 33
§ 3.3
Implicit Differentiation and Related Rates

22. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 22 of 33
 Implicit Differentiation
 General Power Rule for Implicit Differentiation
 Related Rates
Section Outline

23. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 23 of 33
Implicit Differentiation

24. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 24 of 33
Implicit Differentiation
EXAMPLE
SOLUTION
Use implicit differentiation to determine the slope of the graph at the given
point.
1
,
3
;
5
4 2
3




 y
x
x
y
The second term, x2, has derivative 2x as usual. We think of the first term, 4y3,
as having the form 4[g(x)]3. To differentiate we use the chain rule:
 
   
   
x
g
x
g
x
g
dx
d


2
3
12
4
or, equivalently,
  .
12
4 2
3
dx
dy
y
y
dx
d


25. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 25 of 33
Implicit Differentiation
On the right side of the original equation, the derivative of the constant
function -5 is zero. Thus implicit differentiation of yields
.
6
12
2
2
2
y
x
y
x
dx
dy


Solving for we have
.
0
2
12 2

 x
dx
dy
y
CONTINUED
5
4 2
3


 x
y
dx
dy
At the point (3, 1) the slope is
 
 
.
2
1
6
3
1
6
3
6 2
1
3
2
1
3








y
x
y
x y
x
dx
dy

26. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 26 of 33
Implicit Differentiation
This is the general power rule for implicit differentiation.

27. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 27 of 33
Implicit Differentiation
EXAMPLE
SOLUTION
Use implicit differentiation to determine
1
4
4
2


 y
xy
x
This is the given equation.
.
dx
dy
Differentiate.
   
1
4
4
2
dx
d
y
xy
x
dx
d



Eliminate the parentheses.
       
1
4
4
2
dx
d
y
dx
d
xy
dx
d
x
dx
d



Differentiate all but the second
term.
  0
4
4
2 


dx
dy
xy
dx
d
x
1
4
4
2


 y
xy
x

28. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 28 of 33
Implicit Differentiation
Use the product rule on the
second term where f (x) = 4x
and g(x) = y.
    0
4
4
4
2 









dx
dy
x
dx
d
y
y
dx
d
x
x
CONTINUED
Differentiate.
0
4
4
4
2 



dx
dy
y
dx
dy
x
x
Subtract so that the terms not
containing dy/dx are on one
side.
y
x
dx
dy
dx
dy
x 4
2
4
4 



Factor.
  y
x
dx
dy
x 4
2
4
4 



Divide.
4
4
4
2




x
y
x
dx
dy

29. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 29 of 33
Related Rates

30. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 30 of 33
Related Rates
EXAMPLE
(Related Rates) An airplane flying 390 feet per second at an altitude of 5000
feet flew directly over an observer. The figure below shows the relationship of
the airplane to the observer at a later time.
(a) Find an equation relating x and y.
(b) Find the value of x when y is 13,000.
(c) How fast is the distance from the observer to the airplane changing at the
time when the airplane is 13,000 feet from the observer? That is, what is
at the time when and y = 13,000?
.
dx
dy
390

dt
dx

31. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 31 of 33
Related Rates
SOLUTION
(a) To find an equation relating x and y, we notice that x and y are the lengths of
two sides of a right triangle. Therefore
2
2
2
5000 y
x 

CONTINUED
.
000
,
000
,
25 2
2
y
x 

(b) To find the value of x when y is 13,000, replace y with 13,000.
2
2
000
,
000
,
25 y
x 
 This is the function from part (a).
 2
2
000
,
13
000
,
000
,
25 
 x Replace y with 13,000.
000
,
000
,
169
000
,
000
,
25 2

 x Square.
000
,
000
,
144
2

x Subtract.

32. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 32 of 33
Related Rates
CONTINUED
000
,
12

x Take the square root of both
sides.
(c) To determine how fast the distance from the observer to the airplane is
changing at the time when the airplane is 13,000 feet from the observer, we
wish to determine the rate at which y is changing at this time.
2
2
000
,
000
,
25 y
x 
 This is the function.
   
2
2
000
,
000
,
25 y
dt
d
x
dt
d

 Differentiate with respect to t.
dt
dy
y
dt
dx
x 2
2  Eliminate parentheses.

33. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 33 of 33
Related Rates
CONTINUED
Therefore, the rate at which the distance from the plane to the observer is
changing for the given values is 360 ft/sec.
    
dt
dy
000
,
13
2
390
000
,
12
2  y = 13,000; x = 12,000; .
390

dt
dx
dt
dy
000
,
26
000
,
360
,
9  Simplify.
dt
dy

360 Divide.