1
$\begingroup$

I know that the derivative of $n^x$ is $n^x\times\ln n$ so i tried to show that with the definition of derivative:$$f'\left(x\right)=\dfrac{df}{dx}\left[n^x\right]\text{ for }n\in\mathbb{R}\\{=\lim_{h\rightarrow0}\dfrac{f\left(x+h\right)-f\left(x\right)}{h}}{=\lim_{h\rightarrow0}\frac{n^{x+h}-n^x}{h}}{=\lim_{h\rightarrow0}\frac{n^x\left(n^h-1\right)}{h}}{=n^x\lim_{h\rightarrow0}\frac{n^h-1}{h}}$$ now I can calculate the limit, lets:$$g\left(h\right)=\frac{n^h-1}{h}$$ $$g\left(0\right)=\frac{n^0-1}{0}=\frac{0}{0}$$$$\therefore g(0)=\frac{\dfrac{d}{dh}\left[n^h-1\right]}{\dfrac{d}{dh}\left[h\right]}=\frac{\dfrac{df\left(0\right)}{dh}\left[n^h\right]}{1}=\dfrac{df\left(0\right)}{dh}\left[n^h\right]$$ so in the end i get: $$\dfrac{df}{dx}\left[n^x\right]=n^x\dfrac{df\left(0\right)}{dx}\left[n^x\right]$$ so my question is how can i prove that $$\dfrac{df\left(0\right)}{dx}\left[n^x\right]=\ln n$$

edit:

i got 2 answers that show that using the fact that $\lim_{z \rightarrow 0}\dfrac{e^z-1}{z}=1$, so how can i prove that using the other definitions of e, i know it is definition but how can i show that this e is equal to the e of $\sum_{n=0}^\infty \frac{1}{n!}$?

$\endgroup$
  • 2
    $\begingroup$ If you are using the definition of derivative, it is illogical to use L'Hôpital's rule. $\endgroup$ – Miguel Sep 26 '17 at 9:09
  • $\begingroup$ @Miguel i proved the case of L'Hôpital's rule for $\frac{0}{0}$ so i dont see why not $\endgroup$ – ℋolo Sep 26 '17 at 9:12
  • $\begingroup$ Because if you can compute the derivative to apply L'Hôpital's rule, you already know the derivative, which was your question to begin with? $\endgroup$ – Miguel Sep 26 '17 at 9:15
  • $\begingroup$ @Miguel why? i proved that in case if $\frac{g(x)}{f(x)}=\frac{0}{0}$ than $\frac{g(x)}{f(x)}=\frac{g'(x)}{f'(x)}$, where did i found the derivative of $n^x$ there? i only use it $\endgroup$ – ℋolo Sep 26 '17 at 9:24
1
$\begingroup$

$n^h = \exp((h \log n))$;

$\dfrac{n^h-1}{h} = \dfrac{\exp(h(\log n))-1}{h};$

$z: = h\log n$.

Then:

$\lim_{h \rightarrow 0}\dfrac{\exp(h(\log n))-1}{h} =$

$\lim_{z \rightarrow 0}$ $\log n \dfrac{\exp(z) -1}{z} =$

$\log n ×1= \log n$.

Used: $\lim_{z \rightarrow 0} \dfrac{\exp(z)-1}{z} =1$.

$\endgroup$
  • $\begingroup$ but how can i show that $\lim_{z \rightarrow 0}\dfrac{e^z -1}{z}=1$? $\endgroup$ – ℋolo Sep 26 '17 at 9:52
  • $\begingroup$ Textbook: Exponential series expansion, with remainder, then inequalities.Simpler, infinite series, exp(z) = 1 +z+z^2/2 +....., subtract 1, divide, take the limit, less stringent.Is ok? $\endgroup$ – Peter Szilas Sep 26 '17 at 10:02
  • $\begingroup$ yes, thank you very much $\endgroup$ – ℋolo Sep 26 '17 at 10:04
  • $\begingroup$ Anytime, welcome, hope it helps.The inequality is kind of standard, derivation can be found in textbooks, Peter $\endgroup$ – Peter Szilas Sep 26 '17 at 10:19
0
$\begingroup$

It depends on what you feel you can assume about the function ln(x) and the number e.

See the link below for an approach similar to yours: http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx

$\endgroup$
  • $\begingroup$ But how did they proved that $\frac{e^h-1}{h}$ when h is a limit to 0 equal to 1? $\endgroup$ – ℋolo Sep 26 '17 at 9:30
0
$\begingroup$

If you don't have a definition of the logarithm handy (or suitable properties taken for granted), you cannot obtain the stated result because the logarithm will not appear by magic from the computation.

Assume that the formula $n^x=e^{x \log n}$ is not allowed. Then to define the powers, you can work via rationals

$$n^{p/q}=\sqrt[q]{n^p}$$ and extend to the reals by continuity.

Using this apporach, you obtain

$$\lim_{h\to0}\frac{n^h-1}h=\lim_{m\to\infty}m(\sqrt[m]n-1)$$

and you can take this as a definition of the logarithm.

$$\log n:=\lim_{m\to\infty}m(\sqrt[m]n-1).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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