Derivative of the Inverse Function

Motivation:

                      Let  f  be a function such that  f: A  B .  When we talk about the inverse function of  f  ,  we mean the function with the ability such that when  f  maps an element c in A to an element d in B,  this inverse function can map d in B back to c in A.  Usually, we denote its inverse function as  f^-1. Please do not confuse with the result by putting 1 divided by f .  It is just a notation to indicate the inverse function.  And it is with the property

 

                               for any c in A.

 

                     Of course, not every function has a corresponding inverse function.  For a function  f  to have a corresponding inverse function,  the mapping from A to B under the influence of  f   has to be one-to-one .  The mapping is 1-to-1 means

 

                    for any two elements p, q in A,  if  p  q,  then

 

                                                    f(p)  f(q) ,   f(p) and f(q) are in B.

 

                This requirement does make sense; otherwise,  there is no way to map the element in B back to “one” element in A – it is going to have multiple elements.  Let’s see an example:

 

                              f(x)=x²  where  x is any real number.

 

                  Thus,  f(2)=4 and f(-2)=4 .  When you consider the inverse mapping of  f,  “4” will be mapped back to “2” and “-2”.  However, it is not a mapping for function. The mapping for function requires that it only maps to one element.  So, in this case, the inverse function for f does not exist.

 

                  But if we change the domain of f,  it is different:

 

                             f(x)=x²  where x is positive real number.

                   We only define the function f  on “positive” number.  So,  when you see “4”, you will know that f(2)=4 and you will not consider “-2” . Naturally, you map “4” back to “2”.   With this in mind, we start to check the derivative of the inverse function of a function.

 

 

Derivative of the Inverse Function

                       Let f^-1 be the inverse function of f  and we denote

                                       u=f^-1(x)

 

                       Thus,

                                      f(f^-1(x)) = x

 

                     We want to find out the derivative function of  f^-1(x) with respect to x . Let’s apply chain rule :

 

                                          = 1

                     Then

                                                    =

 

                    It means you can find the derivative of the inverse function via the derivative of  f .  Let’s take a look at the following example.

 

 

 

Example:   Let’s define f on positive numbers only,  f(x)=x² .  Thus,  f^-1(x)= .

 

                                 u= f^-1(x)= .

 

                                  f^’(x)=2x 

 

                         So,   

                                   =  =  = ½ x^-1/2

 

 

                Previously,   we know the derivative of f(x)=xⁿ where n is natural number.  We will extend that to fraction exponent.

 

 

 

 

 

 

Theorem:  Let   f(x)=x^1/p  ,  where p is a natural number and x is in the set that f is well defined.  Then

                                             f’(x)=

 

           Proof:

                                 Let   u= f(x)=x^1/p   .

                              u^p = x

                             pu^p-1  = 1    ( Chain rule )

                               =    

                                     let’s substitute u by x^1/p :

                              =

 

 

  Theorem:  Let  f(x)=x^q/p , where p, q are natural numbers.  Then

                                                  f’(x)=  

 

              Proof:

 

                                       f(x) = (x^1/p)^q

 

                                   Let’s apply chain rule on it:

 

                                        f’(x)=q ( x^1/p)^q-1 (1/p) x^1/p-1

                                               =

 

 

                    So,  up to this moment,  we know how to find the derivative of  x^t  where t is any rational number.   Can we extend it to any real number as long as x in the set that is “well-defined” ?   We might try with the following:

 

                                    y=x^t  ,   t might be irrational number .

                      ln(y) = t ln(x)

                        

                          = t x^t-1  

 

                 It looks like the result still hold for irrational exponent.  However,  we have to be very careful about this:  when we take logarithm both sides, that implies the argument in the logarithm function is positive;   when we see  and ,  that implies they can not be zero.  For those restrictions,  we need to keep in mind along the way when we solve any problems.