# Math Help - Derivative at f1(0)

1. Let me state how I define exponentials.

Define $\ln x = \int_1^x \frac{1}{t} dt \mbox{ for }x>0$

It is not hard to show that,
1) $\ln x$ is an increasing function.
2) $\ln x$ is differenciable and therefore continous for $(0,\infty)$.
3) $(\ln x)' = \frac{1}{x} \mbox{ for }x>0$.
4)There exists a number, called $e$, so that $\ln e = 1$.

We see that $\ln x$ is a one-to-one function. Define $\exp x$ to be its inverse function on its range.

It is not hard to show that $\exp x$ is increasing, differenciable and so continous, and furthermore, $(\exp x)' = \exp x$.

We have the following supprising property that if $q$ is a rational number then $\exp q = e^q$. So we define $e^r$, to be $\exp r$. With that we generalize exponents as follows $a^x = e^{\ln a x} = \exp (\ln a x) \mbox{ for }a>0$.

The difficutly besides for using the definition is to have a formal definition of what an exponent means.

2. Originally Posted by ThePerfectHacker
Let me state how I define exponentials.

Define $\ln x = \int_1^x \frac{1}{t} dt \mbox{ for }x>0$

It is not hard to show that,
1) $\ln x$ is an increasing function.
2) $\ln x$ is differenciable and therefore continous for $(0,\infty)$.
3) $(\ln x)' = \frac{1}{x} \mbox{ for }x>0$.
4)There exists a number, called $e$, so that $\ln e = 1$.

We see that $\ln x$ is a one-to-one function. Define $\exp x$ to be its inverse function on its range.

It is not hard to show that $\exp x$ is increasing, differenciable and so continous, and furthermore, $(\exp x)' = \exp x$.

We have the following supprising property that if $q$ is a rational number then $\exp q = e^q$. So we define $e^r$, to be $\exp r$. With that we generalize exponents as follows $a^x = e^{\ln a x} = \exp (\ln a x) \mbox{ for }a>0$.

The difficutly besides for using the definition is to have a formal definition of what an exponent means.
Your definitions of the natural logerithm are completely valid, but I'm confused by something. Why would you define exp q to be anything other than $e^q$? I thought exp q was simply another notation for $e^q$. Am I mistaken?

3. Originally Posted by ThePerfectHacker
Define $\ln x = \int_1^x \frac{1}{t} dt \mbox{ for }x>0$
This is a standard definition for the logarithm function.
Using u-substitution we prove that: ln(a+b)=ln(a)+ln(b).

4. Originally Posted by Plato
This is a standard definition for the logarithm function.
Using u-substitution we prove that: ln(a+b)=ln(a)+ln(b).
Did you mean ln(ab) = ln(a) + ln(b)?

5. Originally Posted by ecMathGeek
Your definitions of the natural logerithm are completely valid, but I'm confused by something. Why would you define exp q to be anything other than $e^q$? I thought exp q was simply another notation for $e^q$. Am I mistaken?
Okay,

1)We define $\ln x$ as the integral above.

2)Using FTC and other things we develop important properties for $\ln x$ such as: $\ln ab = \ln a + \ln b \mbox{ for }a,b>0$.

3)I said that $\ln x$ is continous and increasing as one of its properties so it has an inverse function which we define as $\exp x$.

Note, #3 does not have anything to do with $e$ from the way it is defined above.

4)It can be shown that if $q$ is a rational number then $\exp q = e^q$ i.e. if $q=a/b$ then $\exp q = (\sqrt[b]{e} )^a$.

5)So we see that the $\exp$ function evaluated at $q$ is the same thing as raising the number $e$ (which is define to be so that $\ln e$) to that power.

6)Now is seems "natural" to write $\exp (x)$ rather as $e^x$ because for rational powers it makes sense, i.e. same as raising exponents. But for irrational powers (though they were still not define) this function still makes sense! Because $\exp x$ is defined for all real $x$.

I think this seems strange to you is that you never learned about what an "exponent" means. Yes $a^n = e\cdot a\cdot ... \cdot a$ $n$-times. That is how we define them for positive integers. [math[a^0=1[/tex] that is a mere definition. But what about $a^q$ where $q$ is rational? You defined it as finding deminator roots. But how do you define $a^r$ where $r$ is any arbitrary number? That takes works. The approach above is the standard and simplest and nicest approach to this problem. Meaning, we define the exponent of $e$ to be $L$ provided that $\ln L = e$.

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My point is that the limit $\lim_{x\to c} \frac{a^x - a^c}{x-c}$ is hard to find because we never appropriately defined what $a^x$ means for $a>0$.

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