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pi · « **on:** December 08, 2010, 08:39:33 pm »

Just going over some of my GMA and methods notes/scraps/scribbles and have got this question:

Why is $\frac{d}{dx}e^x=e^x$ ?

And also (hence) $\int e^x\, dx=e^x+c$ ?

Just a bit confused, as the derivatives/anti-derivatives of say $a^x$ , do not do this...

pi · « **Reply #1 on:** December 09, 2010, 10:32:18 pm »

Got it now, don't worry

The definition $e=\lim_{n \to \infty}(1+\frac{1}{n})^n$ is essential.

From the definintion of the derivative: $f'(x)=\frac{f(x+\Delta x)-f(x)}{\Delta x}$ when $\Delta x \to 0$
$f'(x)=\frac{e^{(x+\Delta x)}-e^x)}{\Delta x}$
$f'(x)=\frac{e^x \times e^{\Delta x}-e^x}{\Delta x}$
$f'(x)=\frac{e^x \times e^{\Delta x - 1}}{\Delta x}$
Now show that $\frac{e^{\Delta x - 1}}{\Delta x} \to 1$ as $\Delta x \to 0$
To do this, one inserts $e={(1+\frac{1}{n})}^n$ when $n \to infinity$
$f'(x)=e^x \times \frac{{(1+\frac{1}{n})}^n \times \Delta x -1}{\Delta x}$
As $n \to infinity$ and $\Delta x \to 0$ , we can get rid of one of the limits by writing $n=\frac{1}{\Delta x}$ . This makes the above equation simplify to
$f'(x)=e^x \times \frac{{(1+\Delta x)}^1 -1}{\Delta x}$
$f'(x)=e^x \times \frac{\Delta x}{\Delta x}$
$f'(x)=e^x$

Andiio · « **Reply #2 on:** December 10, 2010, 09:36:29 pm »

Derivatives of a^x and e^x are different as 'e' is not a variable; it is a naturally occurring number just like pi.

pi · « **Reply #3 on:** December 10, 2010, 09:40:06 pm »

Quote from: Andiio on December 10, 2010, 09:36:29 pm

Derivatives of a^x and e^x are different as 'e' is not a variable; it is a naturally occurring number just like pi.

I know, but just wondering why the derivative of e^x is e^x (derivative of pi^x is not pi^x).

It seemed odd, just wondering why 'e' not any other number.. Comes down to first principles

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