Author Topic: Differentiation of e^x (Read 801 times) Tweet Share

brightsky · « **on:** January 04, 2010, 05:22:10 pm »

$f(x) = e^x$

From first principles:

$f'(x) = \lim_{h\to0} \frac{e^{x+h} 0-e^x}{h}$

$\implies \lim_{h\to0} \frac {e^xe^h - e^x}{h}$

$\implies \lim_{h\to0} \frac {e^x(e^h - 1)}{h}$

$\implies e^x \lim_ {h\to0} \frac {e^h - 1}{h}$

What do I do from here? This is an indeterminate form of type 0/0, which entails us to use L'Hospital's rule. But using this, we'll need to find the derivative of e^h, which defeats the purpose of solving the derivative for e^x. Any ideas on how to go about this?

Edited: Latex issues.

brightsky · « **Reply #1 on:** January 04, 2010, 08:26:40 pm »

NB: The book substitutes values of h close to zero to deduce $\lim_{h\to0} \frac {e^h - 1}{h}$ but I was looking for a better way than this, if there is??

zzdfa · « **Reply #2 on:** January 04, 2010, 08:43:46 pm »

if you want to prove that $\lim_{h\to0} \frac {e^h - 1}{h}=1$ , you need to know what $e$ is first. becomes sometimes that limit is taken as the definition of e. that is, e is defined the unique number such that $\lim_{h\to 0}\frac{e^h-1}{h}=1$

a few more ways of defining e are:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots$

$e^x=\lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n$

with some more advanced maths you can show that all these definitions define the same number i.e.

$2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots=\lim_{n\to \infty} (1+\frac{1}{n})^n=\lim_{h\to 0}\frac{e^h-1}{h}=1$

TrueTears · « **Reply #3 on:** January 04, 2010, 08:44:26 pm »

Quote from: brightsky on January 04, 2010, 08:26:40 pm

NB: The book substitutes values of h close to zero to deduce $\lim_{h\to0} \frac {e^h - 1}{h}$ but I was looking for a better way than this, if there is??

It is just the fundamental limit...

/0 · « **Reply #4 on:** January 04, 2010, 08:59:51 pm »

If you accept the 'compound interest' definition of e:

$e = \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n$

You can use that to prove $\lim_{h\to 0} \frac{e^h-1}{h}=1$

Let $u = e^h-1$ , then $h = \log_e(u+1)$

As $h \to 0$ , $u \to 0$ as well, so the limit is equivalent to:

$\lim_{u \to 0} \frac{u}{\log_e(u+1)}=\lim_{u \to 0} \frac{1}{\frac{1}{u}\log_e(u+1)} = \lim_{u \to 0} \frac{1}{\log_e\left((u+1)^\frac{1}{u}\right)}$

Let $t = \frac{1}{u}$ . As $u \to 0$ , $t \to \infty$ . Then the limit is equivalent to:

$\lim_{t \to \infty} \frac{1}{\log_e\left(\left(\dfrac{1}{t}+1\right)^t\right)}$

Since $\lim_{t \to \infty} \left(\frac{1}{t}+1\right)^t = e$ , we get

$\lim_{t \to \infty}\frac{1}{\log_e e} = \lim_{t \to \infty} 1 = 1$

brightsky · « **Reply #5 on:** January 04, 2010, 11:57:12 pm »

Ahh ok! Thanks!!

One thing, isn't d/dx(e^x) = e^x?

superflya · « **Reply #6 on:** January 04, 2010, 11:59:12 pm »

Ilovemathsmeth · « **Reply #7 on:** January 12, 2010, 12:29:52 am »

You don't need to know derivations for the Methods course

I stressed about understanding them. If someone had told me, "you don't need to know derivations" I'd have probably enjoyed seeing how they were derived. Not sure if I'd remember how to derive things though

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Author Topic: Differentiation of e^x (Read 801 times) Tweet Share

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