# I'm not getting e

• Mar 28th 2008, 11:35 PM
Boris B
I'm not getting e
I finally figured out in the assignment I'm doing that one of the variables, known as e, is the e. I.e., it's not actually a variable.

I gather (but can't prove) that of the properties of e is that"
$f(x) = e^x$
$f'(x) = e^x$

Having forgotten that e is not just a regular variable, I decided to take the derivative the old-fashioned way:
$f(x) = e^x$
$f'(x) = x \cdot e^{x-1}$

This leads me astray. Plugging in x=2, I find that ordinary derivation gives me:
$f(2) = e^2$
$f'(2) = 2 \cdot e^{2-1} = 2e$

and I'm pretty sure

$e^2 = 2e$
only when e=2. And because this is the e, it doesn't equal 2!

I can solve the problem now, only I have somewhat less faith in my new ability (?) to take a basic derivative. Is it just that ordinary derivation doesn't work with transcendental / irrational numbers, or did I derive wrong?
• Mar 28th 2008, 11:49 PM
janvdl
Quote:

Originally Posted by Boris B
I finally figured out in the assignment I'm doing that one of the variables, known as e, is the e. I.e., it's not actually a variable.

I gather (but can't prove) that of the properties of e is that"
$f(x) = e^x$
$f'(x) = e^x$

Having forgotten that e is not just a regular variable, I decided to take the derivative the old-fashioned way:
$f(x) = e^x$
$f'(x) = x \cdot e^{x-1}$

This leads me astray. Plugging in x=2, I find that ordinary derivation gives me:
$f(2) = e^2$
$f'(2) = 2 \cdot e^{2-1} = 2e$

and I'm pretty sure

$e^2 = 2e$
only when e=2. And because this is the e, it doesn't equal 2!

I can solve the problem now, only I have somewhat less faith in my new ability (?) to take a basic derivative. Is it just that ordinary derivation doesn't work with transcendental / irrational numbers, or did I derive wrong?

Recall the chain rule.

$\frac{d}{du} e^{f(x)} = e^{f(x)} \cdot f'(x)$
• Mar 29th 2008, 01:15 AM
Opalg
Quote:

Originally Posted by Boris B
Having forgotten that e is not just a regular variable, I decided to take the derivative the old-fashioned way:
$f(x) = e^x$
$f'(x) = x \cdot e^{x-1}$

The reason it leads you astray is that the "old-fashioned way" doesn't apply here. It only tells you how to differentiate a function of the form $x^a$, where the base is the variable x, and the exponent is the fixed number a. But in the function $e^x$ it's the other way round: the base is the fixed number e, and the exponent is the variable x. So you need a different technique altogether.

Quote:

Originally Posted by Boris B
I gather (but can't prove) that of the properties of e is that
$f(x) = e^x$
$f'(x) = e^x$

That's correct, and it's not surprising that you can't prove it, any more than you could prove the "old-fashioned" rule, unless a good teacher (or a good textbook) showed you how.