Derivative of a^x.

Attachment 29739

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

"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...

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Originally Posted by sepoto

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

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

You need to know $\displaystyle {\lim _{h \to 0}}\frac{{{a^h} - 1}}{h} = \log (a)$

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:

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

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.

You seem to be missing the whole point of a "limit". The only time you can find $\displaystyle \lim_{x\to a} f(x)$ by calculating f(a) is if the function, f, is "continuous at x= a". Because we define "f is continuous at a" by "$\displaystyle \lim_{x\to a} f(x)= f(a)$, 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:

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

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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.