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 - (Ö3 /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  u =  x3,     we get    Functions contents F 