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?
 A: Let $f : \mathbb{R} \to \mathbb{R}$ be given by $f(x)=a^x$ and consider the $\ln$ function. We can take the composition so that we have:
$$(\ln\circ f)(x)=\ln (a^x)=x\ln a$$
Now, if we take the derivative, on the left hand side we use the chain rule and on the right hand side we differentiate as usual so that we have:
$$\frac{f'(x)}{f(x)}=\ln a$$
Now solving for $f'(x)$ gives $f'(x) = f(x) \ln a$ so that $f'(x) = a^x \ln a$. This useful technique can be used to take derivatives of other functions: we compose the original function with the inverse and then differentiate on both sides and use the same idea we've used here, this technique can simplify many derivatives and save a lot of time in some situations.
A: Hint: Write $y = a^x$ or equivalently $\ln y = x \ln a$ and use implicit differentiation.
A: Hint: $a^x=e^{\ln a^x}=e^{x\ln a}$.
A: Here are the steps
$$
\frac{d}{dx} \left[a^x\right] = \frac{d}{dx} \left[e^{\ln a^x}\right]= e^{\ln a^x} \frac{d}{dx} \left[\ln a^x\right]
$$
$$
= a^x \frac{d}{dx} \left[x\ln a\right] = a^x\left(\ln a\right)\frac{d}{dx} \left[x\right]= a^x\ln a
$$
A: , the last term, , can be explained by L'Hopital's rule, just taking the limits of the numerator and denominator one of each time.
A: $ \frac{d}{dx} a^x=$
$=d/dx$ $e^{ln\ (a^x)}=$
$=d/dx \ e^{x\ln(a)}$ just as you said.
Now it's time for chain rule.
$f(x)=e^x$
$g(x)=x*ln(a)$
$a^x= f(g(x))$
last comment before I get solving
$a=e^{ln(a)}$
now let's get it done
$d/dx\ e^{x\ ln(a)}=$
$=e^{x*ln(a)}\ d/dx\ (x\ ln(a))=$  (by chain rule)
$=e^{x\ ln(a)} * ln(a)$ 
and that is the solution. Wait it doesnt look correct --> we should do some algebra! 
Show for yourself that 
$e^{x*ln(a)}=(  e^{ln(a)}  )^x$   but as we said that $a = e^{ln(a)}$ so $(  e^{ln(a)}  )^x$  actually equals $a^x$  and of course $e^{x*ln(a)}$ equals a^x
With what we just said you can see that $e^{x * ln(a)}\ ln(a)$ equals $a^x\ ln(a)$ or
$$d/dx a^x = a^x\ ln(a)$$
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!
