"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."

ebendasqc

ebendasqc

Answered question

2022-10-17

The exponential function is the unique function satisfying
f′=f and f(0)=1
however, 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 = 1
so I feel like I must be missing something special about e x . Any pointers would be greatly appreciated.

Answer & Explanation

occuffick24

occuffick24

Beginner2022-10-18Added 13 answers

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

Alexander Lewis

Beginner2022-10-19Added 7 answers

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

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