# proving the exponential function is differentiable

• Nov 21st 2010, 07:18 AM
cooltowns
proving the exponential function is differentiable
Hi my question is :

Show that $\displaystyle e^{x}$is differentiable on $\displaystyle \mathbb{R}$

I began by the definition that A function$\displaystyle f:A\rightarrow\mathbb{R}$is differentiable at c if and only if

$\displaystyle \underset{x\rightarrow c}{lim}\frac{f(x)-f(c)}{x-c}$

I started by$\displaystyle \underset{x\rightarrow c}{lim}\frac{e^{x}-e^{c}}{x-c}$however i am now unsure on how to cancel the $\displaystyle x-c$ bit.

I am not sure on how to show this.

Thanks
• Nov 21st 2010, 07:31 AM
roninpro
It might be better to consider the alternate definition:

$\displaystyle \displaystyle f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$

Then you would have to compute

$\displaystyle \displaystyle \lim_{h\to 0} \frac{e^{x+h}-e^x}{h}=e^x \lim_{h\to 0} \frac{e^h-1}{h}$

(Handling the remaining limit depends on how you define $\displaystyle e$.)
• Nov 21st 2010, 08:00 AM
chisigma
Quote:

Originally Posted by cooltowns
Hi my question is :

Show that $\displaystyle e^{x}$is differentiable on $\displaystyle \mathbb{R}$

I began by the definition that A function$\displaystyle f:A\rightarrow\mathbb{R}$is differentiable at c if and only if

$\displaystyle \underset{x\rightarrow c}{lim}\frac{f(x)-f(c)}{x-c}$

I started by$\displaystyle \underset{x\rightarrow c}{lim}\frac{e^{x}-e^{c}}{x-c}$however i am now unsure on how to cancel the $\displaystyle x-c$ bit.

I am not sure on how to show this.

Thanks

In...

http://www.mathhelpforum.com/math-he...tml#post582679

... it has been extablished that f(x) is differentiable in x=a if there exist a real constant k such that...

$\displaystyle \displaystyle f(x) = f(a) + k\ (x-a) + o(x-a)$ (1)

Now if You remember the 'fundamental limit'...

$\displaystyle \displaystyle \lim_{x \rightarrow a} \frac{e^{x-a}-1}{x-a} = 1$ (2)

... You from (2) derive that is...

$\displaystyle \displaystyle e^{x-a} -1 = (x-a) + o(x-a) \implies \frac{e^{x}}{e^{a}} = 1+(x-a) + o(x-a) \implies e^{x} = e^{a} + e^{a}\ (x-a) + o(x-a)$ (3)

... so that (1) is verified...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$