# Checking derivation of y = a^x

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.

-
Looks all good to me. –  user157545 Jun 25 '14 at 6:30
If $a=e$, then we have the familiar $y=e^x$ and its derivative $$\frac{d}{dx} e^x=\ln(e) e^x=1\cdot e^x=e^x.$$ –  les guimauves Jun 25 '14 at 6:34
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

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