Compound interest ($r$ in terms of $\frac{dA}{dt}$) 
Suppose the amount of money in bank account that is compounded annually is given by $A(t)$. The annual rate of interest is $r$. Find a relation between $\dfrac{dA}{dt}$ and $r$.

My attempt:
$$A(t)=A(0)(1+r)^t \implies
\displaystyle\frac{dA}{dt}=A(0)(1+r)^t\ln (1+r).$$
Dividing, we get
$$\frac{1}{A}\frac{dA}{dt}=\ln(1+r).$$
However, I think the answer should be 
$$r=\frac{1}{A}\frac{dA}{dt}.$$
What am I doing wrong?
 A: This formula $$ \displaystyle r=\frac{1}{A}\frac{dA}{dt}$$ is for continuous compounded interest.
Note that the above formula, is the same as $$\frac{dA}{dt} = r A(t)$$ which has the exponential solution, $$A(t)=A(0)e^{rt}$$
A: The Question is Ill-Posed
First off, I would try to be a little more careful defining terms.  If interest is compounded annually, this means that one interest payment is made every year.  So, if $0 \le t < 1$, then no interest payments have been made, and
$$ A(t) = A(0)$$
(i.e. the amount of money in the account is equal to the initial investment).  At the end of the first year, an interest payment is made, so
$$ A(1) = A(0) + A(0)r = (1+r) A(0). $$
Indeed, the same applies to any value of $t$ such that $1 \le t < 2$.  This pattern can be extended:
$$
A(t) = \begin{cases}
A(0) & \text{if $0\le t < 1$,} \\
(1+r)A(0) & \text{if $1\le t < 2$,} \\
(1+r)^2 A(0) & \text{if $2\le t < 3$,} \\
(1+r)^3 A(0) & \text{if $3\le t < 4$,} \\
\dotsb \\
(1+r)^n A(0) & \text{if $n \le t < n+1$}.
\end{cases}
$$
Slightly more succinctly,
$$ A(t) = (1+r)^{\lfloor t \rfloor} A(0). $$
Notice that $A$ is a step function.  It is constant "most" of the time, then "jumps" at the end of the year.  The derivative is not defined at each of the jumps, and at all of the other points where the derivative is defined, it is equal to zero.  That is
$$
\begin{cases}\left.\dfrac{\mathrm{d}A}{\mathrm{d}t}\right|_t
= 0 &\text{when $t \not\in\mathbb{N}$, and} \\
&\text{is undefined otherwise.}
\end{cases}$$
In the language of measure theory, the derivative is zero "almost everywhere."  In this sense, the question, as posed, is kind of nonsense.
A Continuous Approximation
That being said, we might try another tack:  we can approximate annually compounded interest by assuming that the interest is compounded continuously.  We know that this is going to give an overestimate (and, frankly, a pretty large overestimate, but the magnitude of the error decreases with the frequency of the compounding, so maybe this is a reasonable approach to take).  Continuously compounded interested is obtained by taking a limit as the compounding frequency increases to infinity.  If we compound the interest $n$ times each year, then
$$ A_n(t) = \left(1+\frac{r}{n}\right)^{nt} A(0).$$
Honestly, we probably should be a little more careful here.  The exponent isn't really $nt$.  It is actually $\lfloor nt \rfloor$.  That is, it should be rounded down to the nearest number of compounding periods which have passed.  However, when we take the limit, this isn't going to matter (by an application of the squeeze theorem).  Hence we end up with
$$ A(t) = \lim_{n\to \infty} A_{n}(t) = \underbrace{\lim_{n\to\infty} \left(1+\frac{r}{n}\right)^{nt}}_{=\mathrm{e}^{rt}} A(0) = \mathrm{e}^{rt} A(0). $$
Then
$$ \left.\frac{\mathrm{d}A}{\mathrm{d}t}\right|_t
=\frac{\mathrm{d}}{\mathrm{d}t} \mathrm{e}^{rt} A(0)
= r \underbrace{\mathrm{e}^{rt} A(0)}_{= A(t)}
= r A(t).$$
Therefore
$$ \frac{\mathrm{d}A}{\mathrm{d}t} = rA, $$
which gives a relation between the derivative (of a continuous approximation of) annually compounded interest and the formula for the amount of money in the account.  Dividing both sides of this relation by $A$ (which is never zero as long as $A(0) \ne 0$), we obtain
$$ r = \frac{1}{A} \frac{\mathrm{d}A}{\mathrm{d}t}, $$
which is the relation which was sought.
