derivative of e^x

Discussion in 'Math' started by S_lannan, Oct 9, 2008.

1. S_lannan Thread Starter Active Member

Jun 20, 2007
247
2
Hi, I suck at math.

Could somebody please, in english explain how the derivative of e^x = e^x.
cheers!

2. Dave Retired Moderator

Nov 17, 2003
6,960
145
If you define e^x as a power series you get:

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

And so on - it is an infinite series. You know that factorial, i.e. !, means that the number is multiplied by all integers down to 1 - so 3! = 3x2x1, 4! = 4x3x2x1, 5! = 5x4x3x2x1, and so on.

Now differentiate each term in the power series:

Diff: 1 = 0

Diff: x = 1

Diff: x^2/(2!) = x (Why? The rule is multiple by the power and make the power 1 less, so x^2/(2!) = x^2/(2x1) which when differentiated gives 2x^/(2!) which simpilifes to x)

Diff: x^3/(3!) = x^2/(2!) (Why? Same reason as above)

Diff: x^4/(4!) = x^3/(3!)

And so on. Note each term when differentiated gives you the same term as the previous term you differentiated. So when you differentiate all the components of the infinte power series for e^x you get the same power series that you started with, i.e. d/dx(e^x) = e^x

Dave

3. Mark44 Well-Known Member

Nov 26, 2007
626
1
Another approach is to consider the general exponential function, f(x) = a^x, and to take the limit, as h --> 0, of (f(x + h) - f(x))/h. Let's also stipulate that the base, a, is positive and not equal to 1.

In the following, all limits are as h --> 0.
lim $\frac{a^{(x +h)} - a^{x}}{h}$
= lim $\frac{a^{x} * a^{h} - a^{x}}{h}$
= lim $\frac{a^{x} * (a^{h} - 1)}{h}$
= a^x * lim $\frac{a^{h} - 1}{h}$ [1]

Now, we're stuck. About the only thing we can say is that whatever we have, if the limit [1] exists, it is the derivative of a^x. IOW, it is f'(x).

We can choose values of a and x, and get approximate values for (a^h - 1)/h.

Set a = 2 and x = 0.
If h = .1, we get ~ 0.71773
If h = .01, we get ~ 0.69555
If h = .001, we get ~ 0.69339
If h = .0001, we get ~0.693171
If h = .00001, we get ~0.69314

It appears that if f(x) = 2^x, f'(0) ≈ 0.693.

If we set a = 3 and x = 0, here are approximate values for the limit [1] for various values of h.
h = .1, we get ~1.16123
h = .01, we get ~1.10467
If you continue this process a few more times, I think you'll be convinced that the limit is approximately 1.0986.
This time, if f(x) = 3^x, then f'(0) ≈ 1.0986.

Since a^x is an increasing function, it must be that for some number between 2 and 3, the derivative of that exponential function, evaluated at 0, must be exactly equal to 1.

So (drum roll, please), let's define e to be exactly that number, which turns out to be about 2.718281828...

Why e? To honor Leonhard Euler, considered to be one of the greatest mathematician of all time.

What we have now is that if f(x) = e^x, f'(0) = 1, so we can tackle the original limit to get f'(x).
As before, all limits are as h --> 0.
lim $\frac{e^{(x +h)} - e^{x}}{h}$
= lim $\frac{e^{x} * e^{h} - e^{x}}{h}$
= lim $\frac{e^{x} * (e^{h} - 1)}{h}$
= e^x * lim $\frac{e^{h} - 1}{h}$ [2]

The limit on the right is nothing more than the derivative of e^x, evaluated at x = 0, and this we have already determined is 1.

Therefore, by the definition of the derivative, and a little finessing, we can finally say that d/dx(e^x) = e^x.

Mark

4. S_lannan Thread Starter Active Member

Jun 20, 2007
247
2
thanks for the explanation guys.
I'll try to drill it into my brain.