Show that $d/dx (a^x) = a^x\ln a$. Show that
$$
\frac{d}{dx} a^x = a^x \ln a.
$$

How would I do a proof for this.  I can't seem to get it to work anyway I try.

I know that
$$
\frac{d}{dx} e^x = e^x.
$$
Does that help me here?
Hint: $a^x=e^{\ln a^x}=e^{x\ln a}$.
Let $y = a^x$

Then taking log on both side to the base $e$

We have;

$ \ln y = \ln a^x$

$ \ln y = x\cdot  \ln a$

Taking derivative with respect to $x$;

$\frac d{dx} \ln y = \frac d{dx} x\cdot \ln a$

(I have applied chain rule here>>)

$ \frac 1y \frac{dy}{dx} = \ln a + 0$

$ \frac {dy}{dx} = y\cdot  \ln a$

We know $y= a^x$

So, $\frac {dy}{dx} = a^x \ln a$

Hope it was easier than other methodologies used to derive it!