# Force of Interest Compounding at Annual Rate i

I'm working through some actuarial practice and am lost as to what's going on with the differentiation here (it's been a while since I've had calc):

Derive an expression for $\delta_t$ if accumulation is based on:

(a) simple interest at annual rate i, and

(b) compound interest at annual rate i.

I understand the answer to (a), which is $\delta_t$ = $\frac{ A'(t)}{A(t)}$ = $\frac{i}{1+i \cdot t}$.

However, I don't understand their method of getting $\ln(1+i)$ for part (b).

Any help would be much appreciated!

(p.s. this is Example 1.13 p.40 from Broverman's Mathematics of Investment and Credit 5th ed)

• If this is just as simple as the differentiation rule for exponential functions, then I think I've got it... I was making it harder than it needed to be because of the $A(0)$ term... – Anton Rasmussen Sep 1 '15 at 2:27

Compound interest is $A(t) = (1+i)^t$. So then $\delta_t = \frac{d}{dt}ln(A(t)) = \frac{d}{dt}t*ln(1+i) = ln(1+i)$
From rewriting $$A(t)=(1+i)^t = e^{t\ln(1+i)}$$ we get by the Chain Rule that $$\delta_t = \frac{A'(t)}{A(t)}=\frac{\ln(1+i)\cdot e^{t\ln(1+i)}}{A(t)}=\ln(1+i).$$