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I've tried for a while myself from first principles and applying various rules, but always end up going in circles. I've gotten as far as

$$ y = a^x $$ $$ \frac{dy}{dx} = a^x \left( \lim_{x \rightarrow 0} \frac{a^h-1}{h} \right) $$

but I have no idea how I should go about cancelling the $h$ in the denominator. Any help is appreciated.

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  • $\begingroup$ Do you know this relation ? $$a^x=\exp(x\ln(a))$$ $\endgroup$ – Fabien Jun 18 '14 at 11:12
  • $\begingroup$ Here it's derived: math.com/tables/derivatives/more/b%5Ex.htm (Using an logarithm/exponential-Identity) $\endgroup$ – Mario Krenn Jun 18 '14 at 11:13
  • $\begingroup$ @NicoDean Interesting, do you have a proof for that identity? $\endgroup$ – user157789 Jun 18 '14 at 11:15
  • $\begingroup$ Actually, you find the answere already at this page: math.stackexchange.com/questions/587275/… $\endgroup$ – Mario Krenn Jun 18 '14 at 11:17
  • $\begingroup$ @user157789 Some people would take it as the definition of $a^x$. $\endgroup$ – Jack M Jun 18 '14 at 11:33
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Using the chain rule and assuming you already know that $(\exp)^\prime=\exp,$ you have:

\begin{align} \frac{\mathrm d}{\mathrm dx}a^x=\frac{\mathrm d}{\mathrm dx}e^{x\ln a}=\ln a\cdot e^{x\ln a}=\ln a\cdot a^x. \end{align}

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Let $y=a^x$. Take the natural logarithm of both sides and rewrite the right-hand side using rules of logarithms to obtain $ln(y)=x\cdot ln(a)$. Differentiate both sides implicitly with respect to $x$. So ${1\over y}y'=ln(a)$. Now multiply both sides of the equation by $y=a^x$ and we have $y'=a^x\cdot ln(a)$.

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