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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.

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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.

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    $\begingroup$ Got it, thanks. $\endgroup$ – Alec Rhea Mar 26 '16 at 0:19
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You've made a mistake. Distribute $x $

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

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

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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.

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

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    $\begingroup$ 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). $\endgroup$ – YoTengoUnLCD Mar 26 '16 at 2:31
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    $\begingroup$ 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 $0^0 = 0^{1-1}=0^1/0^1 = 0/0$. $\endgroup$ – mathguy Mar 26 '16 at 3:00
  • $\begingroup$ The problem is that most simplistic, power algorithms make the shortcut of $x^0$ == 1, but clearly $0^0$ == exp(0*log(0)) == NaN. $\endgroup$ – dhchdhd Mar 26 '16 at 3:21
  • $\begingroup$ 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. $\endgroup$ – mathguy Mar 26 '16 at 3:28
  • $\begingroup$ @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 $0^0=1$. $\endgroup$ – Richard Mar 26 '16 at 9:13

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