<p>You are not doing anything wrong. Since
$$
\lim_{r \to 0} \frac{\ln(1+r)}{r} = 1
$$
(to see why: use <a href="https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule" rel="nofollow noreferrer">l'Hôpital's rule</a> or Taylor approximation) holds, for small numbers $r$ 
$$
\ln(1+r) \approx r.
$$</p>

CC BY-SA 4.0
2018-05-24T21:56:58.877Z
2018-05-24T22:17:56.833Z
2018-05-24T22:17:56.833Z
Giskard
<p>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 $\displaystyle\frac{dA}{dt}$ and $r$.</p>

<p><strong>My attempt:</strong></p>

<p>$A(t)=A(0)(1+r)^t$</p>

<p>$\displaystyle\frac{dA}{dt}=A(0)(1+r)^t\ln (1+r)$</p>

<p>Dividing, we get</p>

<p>$\displaystyle\frac{1}{A}\frac{dA}{dt}=\ln(1+r)$</p>

<p>However, I think the answer should be $\displaystyle r=\frac{1}{A}\frac{dA}{dt}$.</p>

<p>What am I doing wrong?</p>

<financial-economics>
Compound interest ($r$ in terms of $\frac{dA}{dt}$)
Siddhartha
# Compound interest ($r$ in terms of $\frac{dA}{dt}$)

(Asked by: username_0 on 2018-05-24)

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 $\displaystyle\frac{dA}{dt}$ and $r$.

**My attempt:**

$A(t)=A(0)(1+r)^t$

$\displaystyle\frac{dA}{dt}=A(0)(1+r)^t\ln (1+r)$

Dividing, we get

$\displaystyle\frac{1}{A}\frac{dA}{dt}=\ln(1+r)$

However, I think the answer should be $\displaystyle r=\frac{1}{A}\frac{dA}{dt}$.

What am I doing wrong?

**Tags:** financial-economics

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## 1 answers

### Answer by username_1 on 2018-05-24. Score: 2 | Accepted answer

You are not doing anything wrong. Since
$$
\lim\_{r \to 0} \frac{\ln(1+r)}{r} = 1
$$
(to see why: use [l'Hôpital's rule](https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule) or Taylor approximation) holds, for small numbers $r$
$$
\ln(1+r) \approx r.
$$


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