# Thread: Derivative of exponential function

1. ## Derivative of exponential function

Hi, we've recently gone over how to get the derivative of exponential and logarithmic functions, except I couldn't make it to that class and we've already moved on. I have the problem as follows:

If $10,000 is invested in a savings account offering 4% per year, compounded semiannually, how fast is the balance growing after 3 years? Things I know: The answer to this problem is$446.02 per year, in the back of the book.

The compound interest formula f(t) = P(1 + r/m)^mt is used.
I have to get the derivative of the formula, and then just plug in the values.

Problem is, I'm unsure how to get the derivative of the formula. I've done all I know how and I'm not getting the answer in the back.

When I tried to find the derivative this is what I came up with, please correct any errors I have made.

f '(t) = (P(1 + r/m)^mt) * ln(mt)

Any help much appreciated.

2. Your f'(t) is not correct.

f(t) = P(1 + r/m)^mt

What are variables here?
Only f(t) and t.
The P, r and m are all constants ---they don't change in the given f(t).

So the function is in the form
f(t) = P[(a)^(g(t))]

d/dx a^u = ln(a) *a^u *du/dx --------------***

Hence,
f(t) = P(1 + r/m)^mt
f'(t) = P[ln(1 +r/m)]*[(1 +r/m)^mt]*m -----(i)

Given:
P = $10,000 r = 4% per annum = 0.04 m = 2 ----semi-annualy t = 3 years Plug those into (i), f'(t) = ($10,000)[ln(1 +0.04/2)]*[(1 +0.04/2)^(2*3)]*2
f'(t) = $446.02 3. Hello, leviathanwave! You missed this differentiation formula: . $\frac{d}{dx}\left(a^u\right) \:=\:a^u\cdot\frac{du}{dx}\cdot\ln a$ In baby-talk, the derivative is: . . (the original function) × (derivative of the exponent) × (natural log of the base)$10,000 is invested in a savings account offering 4% per year, compounded semiannually.
How fast is the balance growing after 3 years?

Things I know:
The answer to this problem is \$446.02 per year, in the back of the book.

The compound interest formula $f(t) \:= \:P\left(1 + \frac{r}{m}\right)^mt$ is used.
I have to get the derivative of the formula, and then just plug in the values. . . . . no
We are given: . $P = 10,000,\; r = 0.04,\;m = 2$

Why not plug them in now? . $f(t) \:=\:10,000\left(1 + \frac{0.04}{2}\right)^{2t}$
. . And our function is: . $f(t) \:=\:10,000(1.02)^{2t}$

Differentiate: . $\frac{df}{dt}\;=\;10,000(1.02)^{2t}\cdot 2\cdot\ln(1.02)$

Now evaluate at $t = 3.$