# Derivative of exp(x^n) by limit definition by waabu

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Derivative of e x by the Limit Definition
Revised—2nd ed.
1

╬
Francis J. O’Brien, Jr., Ph.D.
Aquidneck Indian Council
Newport, RI
November 27, 2012

Preliminaries

By the limit definition of a derivative,

df ( x)       f ( x  h)  f  x 
 lim
dx      h 0          h
n
or, for the function f ( x)  e x ,

ex  h   e x
n                  n       n
de x
 lim                  .
dx    h 0          h

For simplicity we assume n is a positive integer.
Different proofs can be given of the solution—by the Chain Rule, implicit differentiation
n
de x                n
of inverse functions, or L’Hôpital’s Rule on factored expressions;             nx n 1e x . We seek
dx
to provide a plausible elementary proof outline by the above limit definition which is not
available in standard texts. The proof of this derivative can be done by the tools of the limit for
e, the infinite binomial (or exponential) series, and the derivative of x n . Convergence proofs for
the binomial and exponential series are outlined in Appendix I & II.
First, we want to express each term as limits in terms of e. We start with the known limit
(which can be verified by applying L’Hôpital’s Rule for indeterminant forms by derivatives),
1

e x  lim1  hx  or, for x =1,
h

h 0

1
I thank Professor Michael Henle, editor of The College Mathematics Journal, for a helpful criticism of an earlier
version.

1

e  lim1  h 
h

h 0

Then, by the laws of exponents, we can express each term f ( x  h) and f (x) of the
derivative in terms of e.

( x  h) n

f ( x  h)  e x  h   lim 1  h 
n                           h
.
h 0
Likewise,
xn
n
f ( x)  e x  lim 1  h 
h
.
h 0

To show the logic of the complete derivation two examples are given for the derivatives
2
of e x and e x .

Example for n = 1 and n = 2

If n  1 , then the derivative is,
de x       ex  h   e x
 lim                 ,
dx h 0          h

and using the above expressions for f ( x  h) and f (x) with n =1, and limit rule for quotients,

xh                         x

lim 1  h                  lim 1  h 
h                          h
de x       h 0                          h 0
 lim                                                       .
dx h 0                               h

This analysis is a summary of several elementary limit calculations.
We can factor the terms of the limit by the following algebraic identity,


a p  a q  a q a p q  1 ,    

xh
where a  1  h  and p & q are the respective exponents of 1  h ,
x
and . Then, by the
h      h
Product Rule for limits,


x  h   x      

x
                            
de x
 lim 1  h 
h
lim 
1  h        h
 1
.
dx   h0                 h  0              h             
                            
                            
                            

But, as provided above, the first limit term is defined as,

x

lim 1  h 
h
 ex ,
h 0
and this gives,

de x            1  h   1           h
 e x lim                   x             x
  e lim  h   e .
dx       h  0      h            h  0 

de x
Thus,       e x , one of the most remarkable functions in mathematics, named in honor of Euler.
dx

This approach provides a general procedure useful for higher powers of the exponential function.

de x
NOTE: What is             ? We rebuild the limit terms by setting x to  x, and constructing the
dx
f ( x  h) and f ( x) terms of the limit definition.

1

e  x  lim 1  hx 
h
h 0
1

e 1  lim 1  h 
h
h 0
x

 
e  x  e 1  lim 1  h   f ( x)
x

h 0
h

xh

 
e  ( x  h)  e 1
xh
 lim 1  h 
h 0
h
 f ( x  h).

d x
The derivative of      e is given as:
dx

de  x       f ( x  h)  f ( x )
 lim
dx     h 0            h
de  x       e  ( x  h)  e  x
 lim
dx     h 0           h
xh                  x

lim 1  h          lim 1  h 
h                  h
x
 lim h 0                    h 0
de
dx     h0                       h
          xh x    
x
                    
de  x                            1  h 
h
 1
 lim 1  h 
h
lim
dx     h 0                h 0            h        
                    

                    

de  x              (1  h)  1
 e  x lim             
dx           h 0       h     
de  x              h
 e  x lim    e  x
dx           h 0  h 

de x
n

This demonstration can be applied to the general case as well,       .
dx

—————————

If n  2 , then, following the logic of above,

ex  h   e x
2                   2      2
de x
 lim
dx    h0          h

 x  h 2                    x2

lim 1  h                           lim 1  h 
h                          h
2
de x       h 0                                   h 0
 lim
dx  h 0                                   h

          x  h 2  x 2 
x2                                   
x2                                                h         
de
 lim 1  h 
h
lim 
1  h                  1
dx      h0               h  0                   h           
                            
                            
                            
           x  h 2  x2         
                                  
2                                h              
de x
 ex
2
lim 
1  h                      1
dx           h  0                 h                
                                  
                                  
                                  

and simplifying,
2
de x
e x2
lim
1  h h  2 x  1 .
dx        h 0         h

At this point we have to invoke an advanced result for infinite series (see Bers, pp. 484-486 & p.
540) in order to expand 1  h h  2 x and evaluate the derivative. The expression 1  h h  2 x is an
infinite binomial series but it achieves a finite quantity (converges) so long as h  1 and
h  2 x  0 is a real number. See Appendix I.
Expanding 1  h h  2 x by the binomial series2 for several terms,

2      e         x  h 2  x2 1
2
The infinite exponential series can also be used by factoring the first formula, e x lim                                 , and
h 0                 h
 x  h 2  x 2                                                     x
expanding e                     by the exponential series. The infinite series for e converges for all values of x [with
 x  h 2  x 2 substituting for x]. The same steps are used for the binomial series expansion and leads to the same
n
de x
result. This note applies to the general case as well,      . See Appendix I, II for convergence proofs of the
dx
binomial and exponential series.

2                                            h3
1  h h  2 x  1  h  2 x h  h  2 x h  2 x  1 h        h  2 x h  2 x  1(h  2 x  2)      
2!                                               3!

We see that all additional terms in the infinite series will involve an h value which increases. If
we then insert the expansion into the derivative expression we can cancel out denominator h, and
set numerator h terms to 0,

2
de x        2
 e x lim
1  h h  2 x  1
dx           h 0          h
                                          h2                                        h3    
1  h  2 x h  h  2 x h  2 x  1     h  2 x h  2 x  1(h  2 x  2)       1
2                                                 2!                                        3!    
 e x lim                                                                                              
h 0                                                   h
2                          2
 ex        lim (2 x)  2 xe x
h 0

It is seen that dividing by h and applying the limit eliminates all terms except the first,
limh  2 x   2 x.
h0

With these examples as a background we can obtain the result for n. The same steps are
followed.

Derivation for n

1
h
Restating the derivative to solve, and using the limit, e  lim 1  h  , and limit rule for
h 0
quotients,

( x  h) n                      xn

lim 1  h                                       lim 1  h 
h                           h
e x  h   e x
n                 n      n
de x                                  h 0                                               h 0
 lim                     lim                                                                           .
dx     h 0           h         h 0                                               h

Factoring the terms of the limit by the algebraic identity used above for n  2 ,


a p  a q  a q a p q  1 ,       
where a  1  h  and p & q are the respective exponents of 1  h . Then by the Product Rule for
limits,
          x  h n  x n    
n                x n                                  
 1  h 
h
de x                                                      1
 lim 1  h 
h
lim                              .
dx     h0              h 0               h             
                             
                             
But, as provided above, the first limit term is defined to be,

xn

lim 1  h 
h          n
= ex
h 0

Thus,
          ( x  h) n  x n    
                              
xn                              h           
de
e x n lim  1  h                    1 .
dx         h 0                 h             
                              
                              
                              

( x  h) n  x n

Next, we expand the term 1  h 
h
by the infinite binomial series3, (1  h) , as
explained earlier, where

( x  h) n  x n
                     .
h

3
The infinite exponential series can also be used as explained above by using the factored expression,
  x  h n  x n 
e
xn
lim
e                1  . This expansion leads to the result:
h0            h         

                    

n       nd                                                       n
d x           x                1 d n         1 d n                x n dx        n 1 x n
e      e  xn                 x ( 0)        x (0) ...  e           nx     e . See Appendix II for convergence
dx                    dx      2! dx          3! dx                     dx
of this exponential series.

This binomial series converges as shown in Appendix I.
Rather than write each term of the exponent  let’s expand infinite series (1  h) for h
and  , and substitute later. By the binomial expansion,

h2                    h3
(1  h)  1  h   (  1)          (  1)(  2)     .
2!                    3!

( x  h) n  x n
Letting 1  h       h            1 be equal to (1  h)  1 ,

                   h2                    h3        
x
n
n     1  h   (  1)      (  1)(  2)       1
de
 e x lim                     2!                    3!        .
dx         h 0                         h                         
                                                   
                                                   

This allows us to cancel out the denominator h, and eliminate the h terms in the
numerator by applying the limit h  0 to each term,
n
de x

dx
                   h2                    h3        
n       1  h   (  1)      (  1)(  2)       1
e x lim                        2!                    3!        
h 0                          h                         
                                                   
                                                   
n                    h                     h2     
e x lim    (  1)   (  1)(  2)            
h 0
2!                    3!     

             n     n
e x n lim   e x n  lim ( x  h)  x   nx n 1e x n .
h 0          h 0

h      


In the next to last step we see all terms are going to 0 in the limit except for the first one.

( x  h) n  x n dx n
We recognize that when  is substituted, the limit, lim                                      nx n1. See Finney
h0        h          dx
n
and Thomas (Chap. 3, “Derivatives,” pp. 142-143) for an elegant proof of dx .
dx

Putting it all together,

n
de x           x  h n  e x n x n dx n               n
 lim e                  e         nx n  1e x
dx      h0           h              dx

NOTE: Working through the same steps outlined for e  x it can be shown that the derivative of
n
e  x is:

n
de  x                n  n                n
  e  x dx  nx n  1e  x .
dx                       dx

To solve, expand the binomial series with  as defined above,

h2                    h3
1  h     1  h   (  1)             (  1)(  2)    
2!                    3!

This results in the last step of the derivation:

n                              n
e x          lim ( )  nx n  1e  x .
h 0

APPENDIX I

x  h n  x n
Convergence of Binomial Expansion (1  h)               h

We reproduce Bers’ derivation of convergence (p. 540) for the binomial series 1  x 
x  h

by the Ratio Test. Based on that calculation, we substitute the quantities     x  h n  x n .
 
           h
Instead of using n as the limiting value as Bers does, we use k since the
x  h n  x n
expansion (1  h)         h           employs n, where n is assumed to be some integer constant,
n  1,2,3,

 
First, we define the k th term in the expansion 1  x  : ak    x k . The k  1st term
k 
 
   k 1                                                      a
 k  1 x
is: ak 1            . Then, by the Ratio Test calculation,   lim k 1 :
                                                         k   ak

a k 1
  lim
k   ak

   k 1
 k  1 x
      
 lim       
k      k
 x
k 
 
!
x k 1
 lim
k  1!  k  1!
k            !
xk
k!  k !

 lim
k!        k ! x
k   k  1!   k  1!

k!   k   k  1!
 lim                                 x
k   k  1k!   k  1!
 lim
  k  x
k   k  1

 
  1
 lim       x
k
k       1
1
k

 lim
0  1 x    lim  x  x
k        1          k 

Since the Ratio Test requires   1 for convergence, then if x <1, the series 1  x 
will converge.
x  hn  x n                   h  x

For the series, (1  h)         h          , we substitute   x  h n  x n    and arrive at the final
                  
         h
    
  1
step, lim       x:
k
k      1
1
k

x  h n  x n  1
  lim         hk              h
k             1
1
k
  x  h n  x n 1 
                    1
         h        k 
 lim                       h
k               1

 lim (0  1)h   h  h
k 

We treat
x  h n  x n    as a constant and find, lim
1
 0.
h                                          k  k

x  h n  x n
Thus, the series (1  h)           h           converges if h <1 by the criteria of the Ratio Test.

APPENDIX II

Convergence of Exponential Expansion e x  h   x
n   n

x 2 x3      xk    x k 1
The infinite exponential series e x  1  x                        is known to
2! 3!       k! k  1!
converge for all values of x. Convergence is demonstrated by the Ratio Test:

x k 1                x
a                      (k  1)!
  lim k 1  lim                         lim       k      0.
k   ak   k                xk        k 
1
1
k!                     k

For expansion, e  x  h   x , we substitute x  h n  xn for x in the above calculation.
n   n

The algebra of the calculation is similar to the binomial series. This leads to:

x  h
n
 xn    
k 1
x  h n  x n
(k  1)!                           k                  0
 lim                                0.
x  h            
lim
k                         k     k               1              1
n
 xn                      1
k
k!

We treat x  h n  xn as a constant and find, lim
1
 0 . Thus, the exponential series
k  k

e x  h   x converges for all values of x  h n  xn , n  1,2,3, .
n   n