The exponential function is the unique function satisfyingf′=f and f(0)=1however, unless I've made a mistake, we have ∂ ∂ x ( a x ) x = x ( a x ) x − 1 a = a x ( a x ) x − 1 = ( a x ) x and ( a 0 ) 0 = 0 0 = 1so I feel like I must be missing something special about e x . Any pointers would be greatly appreciated.

Question

The exponential function is the unique function satisfyingf′=f and f(0)=1however, unless I&#039;ve made a mistake, we have      ∂          ∂      x        (  a  x      )    x    =  x  (  a  x      )          x      −      1        a  =  a  x  (  a  x      )          x      −      1        =  (  a  x      )    x  and  (  a  0      )    0    =      0    0    =  1so I feel like I must be missing something special about       e    x  . Any pointers would be greatly appreciated.

occuffick24 · Accepted Answer

You differentiated       x    x   wrong. In fact,  (      x    x        )    ′    =  (      e          x      log      ⁡      x            )    ′        =    chain rule    [  x  log  ⁡  x      ]    ′        e          x      log      ⁡      x        =  (  log  ⁡  x  +  1  )      x    x  The rule   [      x    n        ]    ′    =  n      x          n      −      1       only applies when n is a fixed constant.

Alexander Lewis · Accepted Answer

You&#039;ve made a mistake. Distribute x  (  a  x      )    x    =      a    x        x    x  Now can what you&#039;ve written for the derivative be true?

"The exponential function is the unique function satisfying f′=f and f(0)=1 however, unless I've made a mistake, we have d/dx(ax)^x=x(ax)^x−1 a=ax(ax)^x−1=(ax)^x and (a0)^0=0^0=1 so I feel like I must be missing something special about ex. Any pointers would be greatly appreciated."

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