# Derivative of any number raised to the power of x

I know this is a piece of elementary calculus but I just worked it out from (more or less) first principles (as I knew what the answer was but did not know why).

Let $y = r^x$ what is $\frac{dy}{dx}$?

$r = e. k$ where $k$ is some constant. Hence $y=(ke)^x$ or $k^xe^x$

$\frac{r}{e} = k$ and so $\frac{e^{ln(r)}}{e} = k$ and so $e^{ln(r) - 1} = k$.

And $e^{x(ln(r) - 1)}e^x = r^x$, so $e^{xln(r)} = r^x$

Applying the chain rule $u = xln(r)$ and $y = e^u$ so $\frac{dy}{du}\frac{du}{dx} = e^u \times ln(r)$

Which is $e^{xln(r)}ln(r) = r^xln(r)$

Now I have written it all out I know there are redundant steps in there, but this is how I (re)discovered it…