209759095-Technique-of-Differentiation-Ppt-03.ppt

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 1of 33
Chapter 3
Techniques of Differentiation

2. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 2of 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 3of 33
§3.1
The Product and Quotient Rules

4. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 4of 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 5of 33
The Product Rule

6. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 6of 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
22
33xx  3
2
xxf  
10
2
3xxg          333333
2
10
2
10
22
10
22
 x
dx
d
xx
dx
d
xxx
dx
d     xxx
dx
d
xx 2333103
10
22
9
22
     .2323103
10
2
9
22
xxxxx 

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

8. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 8of 33
The Quotient Rule
EXAMPLE
SOLUTION
Differentiate.
Let and . Then, using the quotient rulex
xx 34
24
  34
24
 xxxf xxg   
2
2424
24 3434
34
x
x
dx
d
xxxx
dx
d
x
x
xx
dx
d







    
2
243
13484
x
xxxxx 
 22
22
2
2
4
2
24
343
343343



 xx
xx
x
x
x
x
xx
Now simplify.

9. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 9of 33
The Quotient Rule
Now let’s differentiate again, but first simplify the expression.xx
x
x
x
x
xx 3434
2424


Now we can differentiate the function in its new form.
CONTINUED13
34

 xxx  
2213
34334

 xxxxx
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 10of 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 rateat 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= t
0). ThustWtLtA 
This is the area function.tLtWtWtLtA 
Differentiate using the product
rule.

11. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 11of 33
Rate of Change
Now, since the width of the rectangle is increasing at a rateof 3 inches per
second, we know W΄(t) = 3. And since the length is increasing at a rateof 4
inches per second, we know L΄(t) = 4.
Now we substitute into the derivative of A.
This is the derivative function.tLtWtWtLtA 
CONTINUED
W΄(t) = 3, L΄(t) = 4, W(t) = 5,
and L(t) = 6.4536
0 tA
Simplify. sec/in 38
2
0tA
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 12of 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 xfxgxgxfxgxf
dx
d

PRODUCT RULE



.
2
xg
xfxgxgxf
xg
xf
dx
d 







QUOTIENT RULE

13. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 13of 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 14of 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 15of 33
The Chain Rule

16. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 16of 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,
24
x
x
xf  
4
1xxg   ,2
4
2
x
x
xf  
3
4xxg  


4
2
4
12
1
4
x
x
xgf 

  

. 412
1
4
34
2
4
xx
x
xgxgfxgf
dx
d













17. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 17of 33
The Chain Rule
EXAMPLE
SOLUTION
Compute using the chain rule.
Since yis not given directly as a function of x, we cannot compute by
differentiating ydirectly with respect to x. We can, however, differentiate with
respect to uthe relation , and getdx
dy 2
2,1 xuuy  dx
dy 1uy .
12
1


udu
dy
Similarly, we can differentiate with respect to xthe relation and get2
2xu .4x
dx
du


18. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 18of 33
The Chain Rule
Applying the chain rule, we obtain.4
12
1
x
udx
du
du
dy
dx
dy



It is usually desirable to express as a function of xalone, so we substitute
2x
2
for uto obtain
CONTINUEDdx
dy .
122
4
2


x
x
dx
dy

19. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 19of 33
Marginal Cost & Time Rate of Change
EXAMPLE
SOLUTION
(Marginal Cost and Time Rate of Change) The cost of manufacturing xcases of
cereal is Cdollars, where . Weekly production at tweeks from
the present is estimated to be x= 6200 + 100tcases..
dx
dC 243  xxC
(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
xx
dx
d
dx
dC 2
3243 
(a) We differentiate C(x).

20. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 20of 33
Marginal Cost & Time Rate of Change
(b) To determine , we use the Chain Rule.dt
dC  1001006200,
2
3  t
dt
d
dt
dx
xdx
dC
Now we rewrite xin terms of tusing x= 6200 + 100t.
CONTINUED100
2
3 






xdt
dx
dx
dC
dt
dC 100
1006200
2
3 







tdt
dC
(c) With respect to time, when t= 2, costs are rising at a rate of
 .per week cereal of cases 5.302100
21006200
2
3
2











tdt
dC

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

22. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 22of 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 23of 33
Implicit Differentiation

24. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 24of 33
Implicit Differentiation
EXAMPLE
SOLUTION
Use implicit differentiation to determine the slope of the graph at the given
point.1,3;54
23
 yxxy
The second term, x
2
, has derivative 2xas usual. We think of the first term, 4y
3
,
as having the form 4[g(x)]
3
. To differentiate we use the chain rule: xgxgxg
dx
d

23
124
or, equivalently, .124
23
dx
dy
yy
dx
d


25. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 25of 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.
612
2
22
y
x
y
x
dx
dy

Solving for we have.0212
2
x
dx
dy
y
CONTINUED54
23
xy dx
dy
At the point (3, 1) the slope is

.
2
1
6
3
16
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 26of 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 27of 33
Implicit Differentiation
EXAMPLE
SOLUTION
Use implicit differentiation to determine144
2
 yxyx
This is the given equation..
dx
dy
Differentiate. 144
2
dx
d
yxyx
dx
d

Eliminate the parentheses. 144
2
dx
d
y
dx
d
xy
dx
d
x
dx
d

Differentiate all but the second
term. 0442 
dx
dy
xy
dx
d
x 144
2
 yxyx

28. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 28of 33
Implicit Differentiation
Use the product rule on the
second term where f(x) = 4x
and g(x) = y.  04442 







dx
dy
x
dx
d
yy
dx
d
xx
CONTINUED
Differentiate.04442 
dx
dy
y
dx
dy
xx
Subtract so that the terms not
containing dy/dxare on one
side.yx
dx
dy
dx
dy
x 4244 
Factor.  yx
dx
dy
x 4244 
Divide.44
42



x
yx
dx
dy

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

30. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 30of 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 xand y.
(b) Find the value of xwhen yis 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 31of 33
Related Rates
SOLUTION
(a) To find an equation relating xand y, we notice that xand yare the lengths of
two sides of a right triangle. Therefore222
5000 yx
CONTINUED.000,000,25
22
yx
(b) To find the value of xwhen yis 13,000, replace ywith 13,000.22
000,000,25 yx
This is the function from part (a). 
22
000,13000,000,25 x
Replace ywith 13,000.000,000,169000,000,25
2
x
Square.000,000,144
2
x
Subtract.

32. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 32of 33
Related Rates
CONTINUED000,12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 yis changing at this time.22
000,000,25 yx
This is the function. 
22
000,000,25 y
dt
d
x
dt
d

Differentiate with respect to t.dt
dy
y
dt
dx
x 22 
Eliminate parentheses.

33. © 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS , 12e–Slide 33of 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
000,132390000,122 
y= 13,000; x= 12,000;.390
dt
dx dt
dy
000,26000,360,9 
Simplify.dt
dy
360
Divide.