# Differential Calculus by niusheng11

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```									               Chapter 6
Differential Calculus

The two basic forms of calculus are
differential calculus and integral calculus.
This chapter will be devoted to the former
and Chapter 7 will be devoted to the
latter. Finally, Chapter 8 will be devoted
to a study of how MATLAB can be used
for calculus operations.

1
Differentiation and the Derivative
The study of calculus usually begins with
the basic definition of a derivative. A
derivative is obtained through the process
of differentiation, and the study of all forms
of differentiation is collectively referred to
as differential calculus.If we begin with a
function and determine its derivative, we
arrive at a new function called the first
derivative. If we differentiate the first
derivative, we arrive at a new function
called the second derivative, and so on.
2
The derivative of a function is the
slope at a given point.
y

y  f ( x)

y

x

x   3
Various Symbols for the Derivative

dy               df ( x)
or f '( x) or
dx                dx

dy       y
Definition:           lim
dx x0 x
4
Figure 6-2(a). Piecewise Linear
Function (Continuous).

(a)   y  f ( x)   Continuous Function

x

5
Figure 6-2(b). Piecewise Linear
Function (Finite Discontinuities).

(b)   y  f ( x)
Discontinuities

x1      x2                     x3       x

6
Piecewise Linear Segment
( x2 , y2 )

y2  y1

( x1 , y1 )
x2  x1
7
Slope of a Piecewise Linear Segment

dy           y2  y1
 slope 
dx           x2  x1

8
Example 6-1. Plot the first derivative
of the function shown below.
(a)   y  f ( x)
12

2     4    6         8   10   x

-12
9
(a)   y  f ( x)
12

2        4          6         8       10   x

-12

(b)
dy
 f '( x )
dx                                       6
3

x

-12                              10
Development of a Simple Derivative

yx      2

y  y  ( x  x)      2

y  y  x  2 xx  (x)
2                       2

11
Development of a Simple Derivative
Continuation

y  2 xx  (x)    2

y
 2 x  x
x
dy        y
 lim      2x
dx x0 x
12
Chain Rule

y  f (u )     u  u ( x)
dy df (u ) du                du
                f '(u )
dx       du dx               dx
df (u )
where   f '(u ) 
du          13
Example 6-2. Approximate the derivative
of y=x2 at x=1 by forming small changes.

y (1)  (1)  1
2

y (1.01)  (1.01)  1.0201
2

y  1.0201  1  0.0201
dy y 0.0201
          2.01
dx x   0.01
14
Example 6-3. The derivative of sin u
with respect to u is given below.
d
 sin u   cos u
du
Use the chain rule to find the
derivative with respect to x of

y  4sin x   2

15
Example 6-3. Continuation.

ux    2

du
 2x
dx
dy           du dy du
 f '(u )   
dx           dx du dx
 4(cos u )(2 x)  8 x cos x 2

16
Table 6-1. Derivatives
f ( x)                       f '( x)          Derivative Number
af ( x)                       af '( x)               D-1
u ( x)  v( x)               u '( x)  v '( x)           D-2
f (u )                      du df (u ) du           D-3
f '(u )      
dx        du dx
a                              0                   D-4
xn             (n  0)                nx n 1                 D-5

un             (n  0)              nu n 1
du
dx                D-6
uv                          dv       du
u       v
dx       dx              D-7
u                         du        dv
v      u                   D-8
v                         dx        dx
2
v
du
eu                         eu
dx                   D-9

17
Table 6-1. Derivatives (Continued)
au                                     du
 ln a  a u                            D-10
dx
ln u                          1 du
D-11
u dx
1 du
log a u
 log a e                              D-12
u dx
sin u
 du 
cos u                                D-13
 dx 
cos u                              du
 sin u                               D-14
dx
tan u                              du
sec 2 u
dx                            D-15

sin 1 u      1     du                   1  
   sin u                   D-16
1  u 2 dx             2          2
cos 1 u      1 du                      0  cos   1
u  
D-17
1  u 2 dx
tan 1 u      1 du                      1  
   tan u                    D-18
1  u 2 dx             2          2
18
Example 6-4. Determine dy/dx for
the function shown below.

y  x sin x
2

dy   dv  du
u v
dx   dx  dx
d  sin x          d x   2

x 2
 sin x
dx                dx
19
Example 6-4. Continuation.

dy
 x cos x  sin x  2 x 
2

dx
 x cos x  2 x sin x
2

20
Example 6-5. Determine dy/dx for
the function shown below.
sin x
y
x
du       dv     d  sin x          d  x
v    u         x              sin x
dy
 dx 2 dx              dx
2
dx
dx        v                        x
x cos x  sin x
          2
x
21
Example 6-6. Determine dy/dx for
the function shown below.
x2                           2
                       x
ye      2                 u
2
 x2 
d  
du      2     1  2x  x
                 
dx      dx      2
x2                      x2
dy                             
e     2
  x    xe     2
dx
22
Higher-Order Derivatives

y  f ( x)
dy             df ( x)
 f '( x) 
dx              dx
2
d y               d f ( x) d  dy 
2
 f ''( x)            
dx 2
dx 2
dx  dx 
3
d y                    3
d f ( x) d   d2y 
3
 f ( x) 
(3)
3
  2
dx               dx      dx  dx 

23
Example 6-7. Determine the 2nd
derivative with respect to x of the
function below.

y  5sin 4 x
dy               d
 5(cos 4 x)  (4 x)  20cos 4 x
dx               dx
2
d y                      d
2
 20   sin 4 x   (4 x)  80sin 4 x
dx                       dx
24
Applications: Maxima and Minima

1. Determine the derivative.
2. Set the derivative to 0 and solve for
values that satisfy the equation.
3. Determine the second derivative.
(a) If second derivative > 0, point is a
minimum.
(b) If second derivative < 0, point is a
maximum.

25
Displacement, Velocity, and Acceleration

Displacement          y
dy
Velocity       v
dt
2
dv d y
Acceleration    a    2
dt dt
26
Example 6-8. Determine local maxima
or minima of function below.

y  f ( x)  x  6 x  9 x  2
3      2

dy
 3x  12 x  9
2

dx
3x  12 x  9  0
2

x  1 and x  3
27
Example 6-8. Continuation.
dy
 3x  12 x  9
2

dx
2
d y
2
 6 x  12
dx
For x = 1, f”(1) = -6. Point is a maximum and
ymax= 6.

For x = 3, f”(3) = 6. Point is a minimum and
ymin = 2.
28

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