# Thread: I'm not getting e

1. ## 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"
$\displaystyle f(x) = e^x$
$\displaystyle f'(x) = e^x$

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

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

and I'm pretty sure

$\displaystyle e^2 = 2e$
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?

2. 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"
$\displaystyle f(x) = e^x$
$\displaystyle f'(x) = e^x$

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

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

and I'm pretty sure

$\displaystyle e^2 = 2e$
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.

$\displaystyle \frac{d}{du} e^{f(x)} = e^{f(x)} \cdot f'(x)$

3. 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:
$\displaystyle f(x) = e^x$
$\displaystyle f'(x) = x \cdot e^{x-1}$

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 $\displaystyle x^a$, where the base is the variable x, and the exponent is the fixed number a. But in the function $\displaystyle e^x$ 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.
$\displaystyle f(x) = e^x$
$\displaystyle f'(x) = e^x$