Derivative of exponential function $\frac{d}{dx}a^x$ I am trying to compute simple derivatives of simple functions, but I got stuck on $\frac{d}{dx}a^x=(\ln{a})a^x$.
I suppose the proof is a simple corollary of $\frac{d}{dx}e^x=e^x$, but I am unable to find it. Can anybody help me?
 A: Hint $a^x=e^{\ln (a^x)} = e^{x \ln a}$
You can also use logarithmic differentiation...
A: Write $a^x$ as: $a^x=e^{\ln a^x}=e^{x\ln a}$ and use the chain rule.
A: Chain rule: $2^x$ is the same as $e^{(\log_e 2)x}$.  To differentiate $(\log_e 2)x$, remember that $\log_e 2$ is a constant.
A: Going about it with limits:
$\frac{d}{dx}\left[a^{x}\right] =\lim_{\Delta x \rightarrow 0}  \frac{a^{x+ \Delta x} - a^{x}}{\Delta x} = \lim_{\Delta x \rightarrow 0}  \frac{a^{x}\left(a^{ \Delta x} - 1\right)}{\Delta x} = a^{x}\lim_{\Delta x \rightarrow 0}  \frac{a^{ \Delta x} - 1}{\Delta x}$
Let $u = a^{\Delta x}$ then $\Delta x = \frac{\ln(u)}{\ln(a)}$ and as $\Delta x \rightarrow 0 $ becomes $u \rightarrow 1$
Hence we arrive at
$\frac{d}{dx}\left[a^{x}\right] =a^{x}\lim_{u \rightarrow 1} \frac{u - 1}{\frac{\ln(u)}{\ln(a)}} = \ln(a)a^{x}\lim_{u \rightarrow 1} \frac{u - 1}{\ln(u)} =\ln(a)a^{x}\frac{0}{0} =\frac{0}{0}$
If you are allowed to use the derivate of the natural logarithm $\frac{d}{du} \ln(u) = \frac{1}{u}$ then employing L'hopitals
$\frac{d}{dx}\left[a^{x}\right] =\ln(a)a^{x}\lim_{u \rightarrow 1} \frac{u - 1}{\ln(u)} =\frac{1}{\frac{1}{u}} = \ln(a)a^{x}\frac{1}{\frac{1}{1}} = \ln(a)a^{x} $
If you aren't able to employ the log derivative you can proof it, it's quite simple actually, you just need to remember that 
$e = \lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^{u} = \lim_{u \rightarrow 0} \left(1 + u\right)^{\frac{1}{u}}$
If you're interested, please comment and I will add. 
Hope this is of help,
Edit - Bugger it, I'll show the proof for the Natural Logarithm
$\frac{d}{dx}\left[\ln(x)\right] = \lim_{\Delta x \rightarrow 0} \frac{\ln(x + \Delta x) - \ln(x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{\ln\left(\frac{x + \Delta x}{x}\right)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{\ln\left(1 + \frac{\Delta x}{x}\right)}{\Delta x}$
Let $u = \frac{\Delta x}{x} \rightarrow \Delta x = ux$. Also $\Delta x \rightarrow 0$ becomes $u \rightarrow 0$ 
Hence we arrive at 
$\frac{d}{dx}\left[\ln(x)\right] = \lim_{\Delta x \rightarrow 0} \frac{\ln\left(1 + \frac{\Delta x}{x}\right)}{\Delta x} = \lim_{u \rightarrow 0}\frac{\ln(1 + u)}{xu} =  \frac{1}{x}\lim_{u \rightarrow 0}\frac{\ln(1 + u)}{u} =   \frac{1}{x}\lim_{u \rightarrow 0}\ln\left[1 + u\right]^{\frac{1}{u}}$
I can't remember the name of the theoerem but if 
$\lim_{x \rightarrow a} f(g(x))$ is continuous then 
$\lim_{x \rightarrow a} f(g(x)) = f\left(\lim_{x \rightarrow a} g(x)\right)$
Hence,
$\frac{d}{dx}\left[\ln(x)\right] = \frac{1}{x}\lim_{u \rightarrow 0}\ln\left[1 + u\right]^{\frac{1}{u}} = \frac{1}{x}\ln\left(\lim_{u \rightarrow 0}\left[1 + u\right]^{\frac{1}{u}} \right)$
From before we know this limit is the transcendental constant $e$
Thus,
$\frac{d}{dx}\left[\ln(x)\right] = \frac{1}{x}\ln\left(\lim_{u \rightarrow 0}\left[1 + u\right]^{\frac{1}{u}} \right) = \frac{1}{x}\ln\left(e \right)  = \frac{1}{x}$
