Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

To my knowledge, the exponential function is the unique function satisfying

$f'=f$ and $f(0)=1$

however, unless I've made a mistake, we have

$$\frac{\partial}{\partial x} (ax)^x = x (ax)^{x-1} a = ax (ax)^{x-1} = (ax)^x$$

and

$$(a0)^0 = 0^0 =1$$

so I feel like I must be missing something special about $e^x$. Any pointers would be greatly appreciated.

share|cite|improve this question
up vote 14 down vote accepted

You differentiated $x^x$ wrong. In fact,

$$ (x^x)' = (e^{x \log x})' \overset{\text{chain rule}}{=} [x \log x]' e^{x\log x} = (\log x +1 )x^x$$

The rule $[x^n]' = n x^{n-1}$ only applies when $n$ is a fixed constant.

share|cite|improve this answer
2  
Got it, thanks. – Alec Rhea Mar 26 at 0:19

You've made a mistake. Distribute $x $

$(ax)^x = a^x x^x $

Now can what you've written for the derivative be true?

share|cite|improve this answer

The rule that $\dfrac d{dx} x^n = nx^{n-1}$ holds when $n$ is constant, i.e. $n$ does not change as $x$ changes. In the case of $(ax)^x$, the exponent changes as $x$ changes, and the power rule is not applicable. You can use logarithmic differentiation in such a case.

share|cite|improve this answer

The simplest of the mistakes in your derivation is $0^0 = 1$. That is incorrect. $0^0$ is undefined (it is the same as $0/0$).

share|cite|improve this answer
2  
I wouldn't call it "the same" as $0/0$. $0^0=1$ is harmless in many cases (read: writing power series in a simpler manner). – YoTengoUnLCD Mar 26 at 2:31
4  
You are free to not call it the same, but $0^0=1$ is never harmless. Or, to put it another way, it is no less harmless than $0/0 = 1$. By "the same" I meant $0^0 = 0^{1-1}=0^1/0^1 = 0/0$. – mathguy Mar 26 at 3:00
    
The problem is that most simplistic, power algorithms make the shortcut of $x^0$ == 1, but clearly $0^0$ == exp(0*log(0)) == NaN. – Barry Mar 26 at 3:21
    
Wow, it never crossed my mind. I just checked, and Oracle (database software) thinks power(0,0) = 1. I hope no one uses that for serious applications... although Lockheed did cost taxpayers (in the U.S.) hundreds of millions of dollars for not using the metric system, so nothing should shock me. – mathguy Mar 26 at 3:28
    
@mathguy Most information I can find indicates that $0^0$ is usually taken to be 1 (i.e. binomial theorem) though technically it is undefined. In this particular case, Calvin Khor's solution also works fine when $0^0=1$. – Richard Mar 26 at 9:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.