Trying to understand the magical number of e?

The derivative of a^x (where a is a constant) is (a^x)ln(a). So for 2^x at x=0, your derivative is (2^0)ln(2) = .693.

Thanks for the reply. I’m going to go back over what I read but that seems different that what I have just been doing with derivatives, could you try and give me a more intuitive sense of why you are timing it by the ln(x)

I’m going to see if I can understand as well

I dont have an intuitive sense of how to work out d/dx N^2, I think that’s what maybe holding me back.

Okay, here’s how I arrived at that derivative:

y = a^x
Take the natural log of both sides. Using the power rule, x goes out front.
ln(y) = x*ln(a)
Now use implicit differentiation (let me know if you want me to explain this step more thoroughly)
1/y dy = ln(a) dx
Move your terms around and
dy/dx = y*ln(a)
But we already established that y = a^x, so…
dy/dx = (a^x)ln(a)

This makes sense because for an exponential equation in which your base is e (a=e), the derivative is just e^x. You do it the same way and get (e^x)ln(e), but as ln(e) = 1, it goes down to just e^x.

Do you want to see a proof of the derivative of n^2, too?

No thats ok

I went through it with no problems I’ll just keep going over it

You can also use the chain rule. For positive a, a^x is exp(x*ln(a)) by definition. (Writing exp() for e^ makes it easier to see whats going on):

(exp(x*ln(a)))’ = exp’(x*ln(a))*(x*ln(a))’.

Since exp is its own derivative and (x*ln(a))’ = ln(a),

(exp(x*ln(a)))’ = exp(x*ln(a))*ln(a) = ln(a)*a^x.

x^2, on the other hand, is x*x. By the product rule,

(x^2)’ = x*x’ + x’*x = x*1 + 1*x = 2x.

Wow. This question is SO out of my field of expertise. I admire you who can think this way enormously!

I can’t help you, but maybe The Derivative Rag can. The “meat” of the video starts in at about 47 or really, the real meat starts at about 1 minute, 10 seconds. Anyway, it’s fun, if nothing else.

and since I’ve been on a Tom Lehrer kick lately, I will also post the Derivative Song. Don’t forget the “carefully.” :-)