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Derivative of e^x

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Date: 10/28/1999 at 03:22:31
From: Reb
Subject: Proof of the derivative of e^x

Hi Doctor Math,

I was wondering if you could tell me how to prove that the derivative
of e^x is e^x. I need a step by step proof.

Thanks a lot.
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Date: 10/28/1999 at 05:36:25
From: Doctor Anthony
Subject: Re: Proof of the derivative of e^x

In the 1730's Euler investigated the result of compounding interest
continuously when a sum of money, say, is invested at compound
interest.

If interest is added once a year we have the usual formula for the
amount, A, with principal P, rate of interest r% per annum, and t the
time in years:

A = P(1 + r/100)^t

If interest were added twice a year, we would replace r by r/2 and t
by 2t. The formula would become:

A = P(1 + r/(2x100))^(2t)

If interest were added three times a year, then at the end of t years
A would be:

A = P(1 + r/(3x100))^(3t)

and if we added interest N times a year, then after t years the amount
A would be

A = P[1 + r/(Nx100)]^(Nt)

Now to simplify the working we let r/(100N) = 1/n, so N = nr/100 and

A = P[1 + 1/n]^(nrt/100)

= P[(1 + 1/n)^n]^(rt/100)

We now let n -> infinity and we must see what happens to the
expression (1 + 1/n)^n as n tends to infinity.

Expanding by the binomial theorem

(1 + 1/n)^n = 1 + n(1/n) + n(n-1)/2! (1/n)^2 +
n(n-1)(n-2)/3! (1/n)^3 + ...

Now take the n's in 1/n^2, 1/n^3, ... in the denominators and
distribute one n to each of the terms n, n-1, n-2, ... in the
numerator, getting

1 x (1-1/n) * (1-2/n) * ...

so we now have

(1 + 1/n)^n = 1 + 1 + 1(1-1/n)/2! + 1(1-1/n)(1-2/n)/3! + ...

Now let n -> infinity and the terms 1/n, 2/n, ... all go to zero,
giving

(1 + 1/n)^n = 1 + 1 + 1/2! + 1/3! + 1/4! + ...

and this series converges to the value we now know as e.

If you consider e^x you get

(1 + 1/n)^(nx)

Expand this by the binomial theorem and you have

(1 + 1/n)^(nx) = 1 + (nx)(1/n) + nx(nx-1)/2! (1/n)^2 + ...

and carrying through the same process of putting the n's in the
denominator into each term in the numerator, as described above, you
obtain

e^x = 1 + x + x^2/2! + x^3/3! + ...

and differentiating this we get

d(e^x)/dx = 0 + 1 + 2x/2! + 3x^2/3! + ...

=     1 + x + x^2/2! + x^3/3! + ...

=   e^x

Reverting to our original problem of compounding interest
continuously, the formula for the amount becomes

A = P.e^(rt/100)

You might like to compare the difference between this and compounding
annually.

If P = 5000, r = 8, t = 12 years

Annual compounding gives

A = 5000(1.08)^12 = 12590.85

Continuous compounding gives

A = 5000.e^(96/100) = 13058.48

The difference is not as great as might be expected.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Calculus
High School Calculus
High School Interest

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