Derivative Proof for exponitials

**Zebles** · February 9th 2010, 01:03 PM

Could someone please show me the proof that the derivative of F(x)=a(e)^x is F'(x)=a(e)^x?

Thank you.

**pickslides** · February 9th 2010, 01:13 PM

This site explains what you need

Derivative e^x

also $\frac{d}{dx}ae^x = a\frac{d}{dx}e^x =ae^x$

**Random Variable** · February 9th 2010, 01:30 PM

$\frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}$

$= \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}$

$= \lim _{\Delta x \to 0} \frac{e^{ x}(e^{\Delta x} - 1)}{\Delta x}$

by defintion, $e = \lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^{n}$

let $n = \frac{1}{\Delta x}$

then $e = \lim_{\Delta x \to 0} (1 + \Delta x)^{1 / \Delta x}$

and $e^{\Delta x} = \lim_{\Delta x \to 0} (1 + \Delta x)$

finally $\frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x} \Delta x}{\Delta x} = e^{x}$

**Drexel28** · February 9th 2010, 01:36 PM

Originally Posted by Random Variable

then $e = \lim_{\Delta x \to 0} (1 + \Delta x)^{1 / \Delta x}$

and $e^{\Delta x} = \lim_{\Delta x \to 0} (1 + \Delta x)$

[/tex]

This makes no sense.

**Random Variable** · February 9th 2010, 01:37 PM

Originally Posted by Drexel28

This makes no sense.

What exactly doesn't make sense?

**Drexel28** · February 9th 2010, 01:45 PM

Originally Posted by Random Variable

What exactly doesn't make sense?

What you have said is analogous to saying that if $m=\int_0^1 2^{\frac{1}{x}}dx$ then $m^x=\int_0^1 2\text{ }dx$ . The $\Delta x$ was a dummy variable, it was part of the limit. Outside of the limit it makes no sense.

**Random Variable** · February 9th 2010, 01:57 PM

What if I did something as simple as reordering the steps?

by defintion, $e = \lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^{n}$

let $n = \frac{1}{\Delta x}$

then $e = \lim_{\Delta x \to 0} (1 + \Delta x)^{1 / \Delta x}$

and $e^{\Delta x} = \lim_{\Delta x \to 0} (1 + \Delta x)$

$\frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}$

$= \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}$

$= \lim _{\Delta x \to 0} \frac{e^{ x}(e^{\Delta x} - 1)}{\Delta x} = \lim _{\Delta x \to 0} \frac{e^{x} \Delta x}{\Delta x} = e^{x}$

**Drexel28** · February 9th 2010, 02:00 PM

Originally Posted by Random Variable

What if I did something as simple as reordering the steps?

by defintion, $e = \lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^{n}$

let $n = \frac{1}{\Delta x}$

then $e = \lim_{\Delta x \to 0} (1 + \Delta x)^{1 / \Delta x}$

and $e^{\Delta x} = \lim_{\Delta x \to 0} (1 + \Delta x)$

$\frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}$

$= \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}$

$= \lim _{\Delta x \to 0} \frac{e^{ x}(e^{\Delta x} - 1)}{\Delta x} = \lim _{\Delta x \to 0} \frac{e^{x} \Delta x}{\Delta x} = e^{x}$

Still no cigar.

**Random Variable** · February 9th 2010, 03:10 PM

OK. How about this:

$e = \lim_{n \to 0} (1 + n)^{1 /n}$

so for very small $n$ , $e \approx (1+ n) ^{1/ n}$

and $e^{n} \approx (1 + n)$

which means $\lim_{n \to 0} \frac{e^{n}-1}{n} = \frac{n}{n} = 1$

$\frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}$

$= \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}$

$= e^{x}\lim _{\Delta x \to 0} \frac{e^{\Delta x} - 1}{\Delta x} = e^{x}(1) = e^{x}$

**Drexel28** · February 9th 2010, 03:17 PM

Originally Posted by Random Variable

OK. How about this:

$e = \lim_{n \to 0} (1 + n)^{1 /n}$

so for very small $n$ , $e \approx (1+ n) ^{1/ n}$

and $e^{n} \approx (1 + n)$

which means $\lim_{n \to 0} \frac{e^{n}-1}{n} = \frac{n}{n} = 1$

$\frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}$

$= \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}$

$= e^{x}\lim _{\Delta x \to 0} \frac{e^{\Delta x} - 1}{\Delta x} = e^{x}(1) = e^{x}$

Better

**Zebles** · February 9th 2010, 04:10 PM

Random variables first proof makes quite a bit of sense to me. thanks for the proofs guys.

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