Derivative of a^x.

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November 15th 2013, 02:40 PM
sepoto

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Derivative of a^x.

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$M(a)=\frac{\lim}{\Delta x\rightarrow 0}\frac{a^\Delta ^x-1}{\Delta x}$
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$M(a)=\frac{d}{dx}a^x...at...x=0$

"2. Geometrically, M(a) is the slope of the graph y = ax at x = 0."

My question is what about the other points other than 0 where a^x is equal to 1? Right now I think that the slope at 2 is M(a)a^x however the documents seem to me right now to be saying that M(a) only is the slope at the other points such as 2. I'm trying to clear this up.

Thanks in advance...
November 15th 2013, 03:09 PM
Plato

Re: Derivative of a^x.

Quote:

Originally Posted by sepoto

$M(a)=\frac{\lim}{\Delta x\rightarrow 0}\frac{a^\Delta ^x-1}{\Delta x}$
-----------
$M(a)=\frac{d}{dx}a^x...at...x=0$

${\lim _{h \to 0}}\frac{{{a^{x + h}} - {a^x}}}{h} = {a^x}{\lim _{h \to 0}}\frac{{{a^h} - 1}}{h}$

You need to know ${\lim _{h \to 0}}\frac{{{a^h} - 1}}{h} = \log (a)$
November 15th 2013, 08:16 PM
sepoto

Re: Derivative of a^x.

I see that is correct and if I use my calculators ln function that ln(2)*2^3 gives me the derivative of 2^x at the point x=3 for the function 2^x.

What I'm still trying to understand is that if the limit is for the point 0 it looks like it would result in a divide by zero error if the closest point 2^0 is entered:

$\frac{2^0-1}{0}$

which does not seem to be coming out to:

0.6931

which is the actual derivative of 2^x at x=0.

P.S.

O.K. Since it's the limit so I would have to plug in a value as close to zero as I can without actually being on zero.
November 16th 2013, 07:15 AM
HallsofIvy

Re: Derivative of a^x.

You seem to be missing the whole point of a "limit". The only time you can find $\lim_{x\to a} f(x)$ by calculating f(a) is if the function, f, is "continuous at x= a". Because we define "f is continuous at a" by " $\lim_{x\to a} f(x)= f(a)$ , that would be circular reasoning except that we have other ways of determining if functions are continuous.

However, even in this case we do NOT take a limit by "plug in a value as close to zero as I can without actually being on zero". There is NO non-zero number closest to 0. We do NOT find limits by "plugging in" values. I don't have time to give you a course in limits here so I suggest you review them in your text book. They are crucial to Calculus and not as trivial as you seem to think.

From your first post:

Quote:

My question is what about the other points other than 0 where a^x is equal to 1?

Unless a= 1, there is NO point other than 0 where a^0= 1. If a= 1 that is true for all x so f(x)= 1^x is precisely the constant function, f(x)= 1 which has derivative f'(x)= 0.