# Math Help - Derivative Proof for exponitials

1. ## Derivative Proof for exponitials

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

Thank you.

2. This site explains what you need

Derivative e^x

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

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

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

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

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

4. Originally Posted by Random Variable

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

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

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This makes no sense.

5. Originally Posted by Drexel28
This makes no sense.
What exactly doesn't make sense?

6. 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.

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

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

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

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

8. 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.

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

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

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

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

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

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

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

10. Originally Posted by Random Variable

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

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

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