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Can you tell me if there are any flaws with this derivation of $y = a^x$...

The assumptions are that the derivative $$\frac{d}{dx}e^x = e^x$$ and that the derivative $$\frac{d}{dx}\ln x = \frac{1}{x} = x^-1.$$

$$\begin{array}{rcl} y &=& a^x\\ \ln y &=& \ln a^x\\ \ln y &=& x \ln a\\ x &=& \frac 1{\ln (a) }\ln y\\ \frac {dx}{dy} &=& \frac {1}{\ln (a) } y^-1\\ \frac {dy}{dx} &=& \ln(a)y\\ \frac {d}{dx} a^x &=& \ln(a)a^x \end{array} $$ I find the natural logarithm in this result very odd, a bit random.

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Looks all good to me. – user157545 Jun 25 '14 at 6:30
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If $a=e$, then we have the familiar $y=e^x$ and its derivative $$\frac{d}{dx} e^x=\ln(e) e^x=1\cdot e^x=e^x.$$ – Cookie Jun 25 '14 at 6:34
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Looks good, except... differentiation – Mar Johnson Jun 25 '14 at 7:10
    
A nitpick, use curly brackets like x^{-1} instead of x^-1 as you used... Otherwise very nice! – Joachim Jun 25 '14 at 12:58
up vote 6 down vote accepted

Your work is correct. Notice also that we can differentiate this function by simpler way:

$$y=a^x=e^{\ln(a) x}$$ so using the chain rule $$\frac{dy}{dx}=\ln(a)e^{\ln(a) x}=\ln(a) a^x$$

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Bravo! That is clever. Thanks. – John Jun 25 '14 at 6:46
    
You're welcome. – user63181 Jun 25 '14 at 6:55

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