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e^(e^x) (Posted on 2011-02-08) Difficulty: 3 of 5
Find a formula, involving n, the number of e's present in the below function, for the derivative of :








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Solution Solution | Comment 4 of 5 |

Define
   f(n,x) = e^f(n-1,x)  if n > 0,
          = x           if n = 0.
   Examples:
       f(0,x) = x
       f(1,x) = e^x
       f(2,x) = e^(e^x)
Therefore,
   f'(n,x) = e^[ f(0,x) + ... + f(n-1,x) ]  if n > 0,
           = 1                              if n = 0.
   Examples:
       f'(0,x) = 1
       f'(1,x) = e^x
       f'(2,x) = e^[ x + e^x ]
Proof by mathematical induction on n:
   f'(0,x) = d[x]/dx = 1
   f'(1,x) = d[e^x]/dx = e^x = e^f(0,x)
   Assume that
   f'(k,x) = e^[ f(0,x) + ... + f(k-1,x) ]  for k > 0.
   f'(k+1,x) = d[e^f(k,x)]/dx = e^f(k,x)*d[f(k,x)]/dx
 
             = e^f(k,x)*e^[ f(0,x) + ... + f(k-1,x) ]
             = e^[ f(0,x) + ... + f(k,x) ]
   QED
 

  Posted by Bractals on 2011-02-08 16:17:36
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