Derivative of Composite Functions and Chain Rule

 

Motivation:

                          To find the derivative function for any function f from the basic definition is not a easy task.  If the derivative functions of some basic functions are known,  using those basic functions could generate many other types of functions. If the rules for derivative functions between the new functions and old functions can be grasped, we can just focus our study on those basic functions.

 

 

Theorem:   Let  f, g, h be functions of x .    g(x) and h(x) are differentiable.

(1)       if  f = g+h ,  then  f’(x)=g’(x) + h’(x) .

(2)       if  f=gh , then  f’(x)=g’(x)h(x)+g(x)h’(x) .

(3)       if     where g(x)  0,    then   

                             

 

       Proof:

 

               (1)

                          f(x+t)  f(x) = g(x+t) + h(x+t)  g(x)  h(x)

                                              = (g(x+t)-g(x)) + (h(x+t) h(x))

 

                          So,  

                                         

 

                           The derivative of  g and h exist.  Thus,

                       

                                   

 

              (2)   

                         f(x+t )- f(x) = g(x+t)h(x+t)  g(x)h(x)

                                           = g(x+t)h(x+t)  g(x+t)h(x) + g(x+t)h(x)  g(x)h(x)

                                           = g(x+t)( h(x+t)-h(x) ) + h(x)( g(x+t)-g(x) )

 

                         So,

                                        

                        Thus,

                                        f’(x)=g’(x)h(x)+g(x)h’(x)

 

 

                  (3)

                             f(x+t)-f(x)

                          =  

                          =  

                         =  

 

                           

                         =  

 

                       So,

                                       

 

 

 

 

 

 

Composite Function

                 A composite function is a combination of two functions by feeding the output of first function as the input of the second function.  For example, if we have two functions  g(x)=3x and h(x)=2x+1,  then the composite function   is

 

                                              

 

                   In this case  t=2x+1,  s=3t .  So,  the composite function is

 

                                  (  )(x) = 3t = 3(2x+1) = 6x+3 .        

 

                  We simply write it as  g(h(x)) .

 

                   And let’s check the derivative of the composite function below.

 

 

 

 

 

 

 

Theorem ( Chain Rule )

                    Let   g and  h  be function of x such that the range of  h  is in the domain of g. g and h are differentiable with derivative functions  and  respectively.  If a function  f  is defined by  f(x)=g(h(x)) , then

 

                                               

            Usually, we also denote it as

                                              

 

        Proof:

                           

                       f(x+t)-f(x) = g(h(x+t))  g(h(x))

                                         =  

 

                      =  

 

                      h is differentiable   h is continuous .

                      Thus,  when  t  0,  h(x+t)  h(x) .  If we write

 

                                      z=h(x)  and  h(x+t)=h(x)+s = z+s

 

                      then  when t  0,  z+s  z .    In other words,  s  0 when t  0 .

 

                     So,

 

                                 

                             =  

                             =  

                            =  

                            =  

 

                      Hence,    .

                              

 

 

               In previous section, we know that when   f(x)=xⁿ ,  f’(x)=nx^n-1 .  So, that can cover a lot of functions that are composed by polynomials.

 

 

 

 

Example 1:  f(x)=2x²+3x+1

                    Then   f’(x)=4x+3

 

 

 

 

 

Example 2:  g(x)=6x²  5

                     Then  g’(x)=12x

 

 

Example 3:  h(x)=( 2x²+3x+1)( 6x²  5)

                     

                        From the previous two examples,

                                   h’(x)=(4x+3)(6x² -5 )+(2x²+3x+1)(12x)

                                           = 24x³ + 18x²  20x  15 +

                                              24x³ + 36x²  12x

                                           = 48x³ + 54x²  32x -15

 

 

Example 4:   h(x)= (x²+3x+1)⁵

 

                       Let f(x)=x⁵  and g(x)=x²+3x+1 .

                       h(x) is the composite function of  f(x) and g(x) :  (  )(x) = f(g(x)) .

                      

                         So,

                                      h’(x)=5(x²+3x+1)⁴(2x+3)

 

 

Example 5: