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Differential Calculus

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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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