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In order to find the derivative of a exponential function, on its general form $a^x$ by the definition, I used limits.

$\begin{align*} \frac{d}{dx} a^x & = \lim_{h \to 0} \left [ \frac{a^{x+h}-a^x}{h} \right ]\\ \\ & =\lim_{h \to 0} \left [ \frac{a^x \cdot a^h-a^x}{h} \right ] \\ \\ &=\lim_{h \to 0} \left [ \frac{a^x \cdot (a^h-1)}{h} \right ] \\ \\ &=a^x \cdot \lim_{h \to 0} \left [\frac {a^h-1}{h} \right ] \end{align*}$

I know that this last limit is equal to $\ln(a)$ but how can I prove it by using basic Algebra and Exponential and Logarithms properties? Thanks

  • 0
    $a^h=\exp(\ln a \cdot h)$. Are you able to Taylor expand the exponential function?2011-10-30
  • 0
    Well, can we use the definition but be a bit witty? Perhaps we could use the definition (along with the chain rule) on $\log y = \log (a^x)$?2011-10-30
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    I think it is worth mentioning that some authors (such as E. Landau) define the logarithm that way.2012-05-02
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    I consider this to be a rather tricky issue. Let me explain. Let $a$ be positive real number. Then to \textit{define} $a^x$ algebraically is easy on the rationals, just ask that the standard exponent laws hold. But if you want to define $a^x$ for irrational values, you need to do some analysis. In particular you need to prove that $a^x$ is continuous on the rationals.2012-06-03
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    http://math.stackexchange.com/questions/1828962/is-it-valid-to-write-1-lim-x-rightarrow-0-fracex-1x-frac-lim-x2016-07-04

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