Applications of the derivative
      Differential of a function
         Use of differential to approximate the value of a function
         Rules for differentials
         Differentials of some basic functions
Differential of a function
If given y = f (x) is differentiable function then, the derivative of y or f (x)
as the instantaneous rate of change of y with respect to x, can also be written as 
where, the increment Dx = dx is called differential of the independent variable and dy  is the differential of  y.
Therefore, the differential of a function
dy = f '(x)dx
represent the main part of the increment Dy which is linear concerning the increment Dx = dx, as is shown
in the figure below.
Example:   Find the increment Dy and the differential dy of the function  f (x) = x2 - 4x, and calculate their values if  x = 3 and  Dx = 0.01.
Solution:  Since  Dy = f (x + Dx) - f (x)  then  Dy = (x + Dx)2 - 4(x + Dx) - (x2 - 4x)
Dy = (2x - 4)Dx + Dx2     therefore,    dy = (2x - 4)Dx = (2x - 4)dx
              or         f '(x) = 2x - 4,     dy = f '(x)dx = (2x - 4)dx.
Let calculate the values of  Dy  and  dy  for  x = 3 and  Dx = 0.01,
                                      Dy = (2x - 4)Dx + Dx2 = (2 3 - 4) 0.01 + 0.012 = 0.0201
   and    dy = f '(x)Dx,     dy = (2x - 4)Dx = (2 3 - 4) 0.01 = 0.02. 
Use of differential to approximate the value of a function
If the change of the independent variable (increment) Dx is small then, the differential dy of a function y = f (x) and the increment Dy are approximately equal,
                             Dy dy    that is,     f (x + Dx) - f(x)f '(x)Dx
                                                   or      f (x + Dx)   f(x) + f '(x)Dx.
Example:   Using the differential calculate approximate value of  cos 61.
Solution:  Since we must use radians as units of measurement of angles, let first convert degrees to radians.
Using the formula     
 f (x + Dx)   f (x)f '(x)Dx  and taking,   f (x) = cos where,  x = arc (60 and  Dx = arc (1)
therefore,
cos (x + Dx)  cos x + (cos x)' Dx  where,  xp/3  and  Dxp/180
     or     cos 61  cos (p/3) + ( - sin (p/3)) (p/180) = 1/2 - (3 /2) 0.0174,     cos 61  0.4849.
Example:   Using the differential calculate approximate value of  e 0.15.
Solution:   Using the formula  
f (x + Dx)   f (x)f '(x)Dx  and taking,   f (x) = ex  where,  x = 0  and  Dx = 0.15   
   then,     ex + Dx   ex + (ex)' Dx     that is,        e 0.15   e 0 + e 0 0.15 = 1 + 0.15 = 1.15.
Example:   Approximately, how much will increase the volume of the sphere if its radius  r = 20 cm increases by 2 mm.
Solution:   Using approximation     Dy dy    or     Dyf (x + Dx) - f (x)   f '(x)Dx
                                                     therefore,         DV    V '(r)Dr.
Since    V = (4/3)pr3   then     V '(r) = (4/3)p 3r2 = 4pr2,
and by substituting  r = 20   and   Dr = 0.2  into
DV    V '(r)Dr = 4pr2 Dx   we get    DV    4p 202 0.2 = 1005,3  cm3.
Rules for differentials
Rules for differentials are the same to those for derivatives, such that
  1)   dc = 0,   c is a constant
  2)   dx = Dx,   x is the independent variable
  3)   d(cu) = c du
  4)   d(u v) = du dv
  5)   d(u v) = u dv v du
 6) 
  7)   d f(u) = f ' (u) du.
Differentials of some basic functions
  1)   d un = n un - 1du
  2)   d eu = eu du
  3)   d an = an ln a du
 4) 
  5)   d sin u = cos u du
  6)   d cos u = - sin u du
 7) 
 8) 
 9) 
 10) 
 11) 
 12) 
Example:   To show the use of the formula 2)  d eu = eu du above, let substitute;  a)  u = x,  b)  u = cos x
and   c)  ux3.
Solution:   a) By substituting  u = x,  obtained is  d ex = ex dx
                 b) By substituting  u = cos x,  we get  d e cos x = ecos x d (cos x) =  - e cos x sin x dx
  c) By substituting  ux3,     we get
Functions contents F
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