What is the derivation for the derivative of $a^{t}$ Been driving me nuts.  Can someone prove to me that 
$$\frac{d(a^t)}{dt} = a^t \ln(a)$$
Thank you!
 A: Express $a^t$ as $e^{\ln a^t}$, leading to
$$\frac{d(a^t)}{dt}=\frac{d(e^{\ln a^{t}})}{dt}=\frac{d(e^{t\ln a})}{dt}=(\ln a)(e^{t\ln a})=(\ln a)(e^{\ln a^{t}})=a^t\ln a$$
This works because by taking the $\ln(a^t)$ then raising $e^{\ln a^{t}} =a^{t}$ as such you are not changing the nature of the function.
A: Use the definition(!) $$a^t:=e^{t\ln a}$$
A: You know  that $$\lim_{h \to 0} \frac{a^h-1}{h} = \log{a}$$, then
$$\frac{d}{dt} a^t = \lim_{h \to 0} \frac{a^{t+h}-a^t}{h} = \lim_{h \to 0} a^t\frac{a^h-1}{h} = a^t \log{a}$$
A: \begin{align}
\frac{d(a^x)}{dx} = \lim_{\Delta x\to 0}\frac{\Delta(a^x)}{\Delta x} = \lim_{\Delta x\to 0}\frac{a^{x+\Delta x}-a^x}{\Delta x} & =\lim_{\Delta x\to 0}\left(a^x \frac{a^{\Delta x}-1}{\Delta x}\right) \\[18pt]
& = a^x \lim_{\Delta x\to 0} \frac{a^{\Delta x}-1}{\Delta x}\quad\ldots\ldots
\end{align}
This last step works because $a$ is "constant" and "constant" means not depending on $\Delta x$, i.e. not changing as $\Delta x$ changes.
$$
\cdots\cdots = \left(a^x\cdot \text{constant}\right) \quad \ldots\ldots
$$
and this time "constant" means not depending on $x$.
If $a=e$ then this last "constant" is $1$.  That is what is "natural" about $e$.
To find the "constant" in other cases, use the chain rule and the laws of exponents.  Others have posted the details of that step here.
A: Alternative proof via implicit differentiation:
Let $y=a^t.$
Taking (natural) logs of both sides, yielding: $\ln(y)=\ln(a^t)\iff \ln(y)=t\ln(a)$ (using the power rule for logarithms).
Now, differentiate both sides wrt $t$, as follows: $\underbrace{\frac{1}{y}\frac{dy}{dt}}_{\text{using chain rule}}=\ln(a)\iff\frac{dy}{dt}=y\ln(a)=\underbrace{a^t\ln(a)}_{\text{since }y:=a^t}$.
