# Some basic derivative-integral issues

1. Dec 14, 2005

### Robokapp

It's not really homework, although that is where I encountered this first time.

I had to evaluate the integral from zero to whatever of 3^x.
I realized...while dy/dx of e^x=e^x the dy/dx of any number not equal to e does not follow the same pattern. It has a different formula.

Same for Log base e (the natural log)...so I'm asking, isn't e and any number n constants? why does the derivative or integral of e^x differ from n^x ?

Also, why is log base e (natural log) different in formul than any other log?

I don't understand why a constant won't act like the others. Is it that e was discovered this way purposelly? Or e was already known and it just hapepned to work this way?

I know I'm splitting hairs here, but it would be useful to understnad how they work, in interest of saving brain capacity...
~Robokapp

2. Dec 14, 2005

### incognitO

its because $\log e=1$...

look at it this way....

$$\begin{array}{c}a^x=y(x) \\ \\x \log a=\log y(x) \\ \\ \dfrac{d}{dx}(x \log a)=\dfrac{d}{dx}\log y(x) \\ \\ \log a =\dfrac{y'(x)}{y(x)} \\ \\ y(x) \log a =y'(x)\end{array}$$

wich means

$$\frac{d}{dx}a^x=a^x \log a$$

$e$ is not different from any other constant, just $\log e=1$

Last edited: Dec 14, 2005
3. Dec 14, 2005

### d_leet

The natural log of e equals one, in other words log base e of e is one but thats the same for any number

suppose we have a

then log base a of a is 1because a^1 is a.

For integrating and differentiating a number n^x we know that

n^x = (e^ln n)^x which is the same as e^(x*ln n) and since ln n is a coinstant the integral would be (e^(x*ln n))/(ln n) which is (n^x)/(ln n)
differentiating is similar.

4. Dec 15, 2005

### hypermorphism

e was discovered to be the base of the natural logarithm, where the natural logarithm was found by studying the break in behavior of the integrals $$\int x^n dx$$ when x=-1. The integral was found to behave exactly like a logarithm, and thus the integral $$\int_1^x \frac{dt}{t}$$ was defined to be the natural logarithm with the natural base e, Euler's number.
It is also fascinating to know that the only family of functions whose derivative returns the same function is f(t) = A*et. This property makes e pop up in a ridiculous amount of places (it is as ubiquitous as $\pi$). You can find out more about the history of e in the book "e: the Story of a Number" by Maor.

Last edited: Dec 15, 2005
5. Dec 15, 2005

### Robokapp

Well...talking about that, I'll probably feel like an idiot for stating the obvious, but did anyone notice that the derivative of y=ln(n x) where n is any constant number is always 1/x

I wonder...how does that work? I mean ln (1000x) shouldn't raise/fall at same rate as ln (0.001x) right? but because of Chain Rule they have equal derivatives. The constant goes in front and cancels out with the one inside Ln.

I understnad the math, but how reasonable is it?

6. Dec 16, 2005

### d_leet

Well if you remember the properties of logarithms specifically this one

log(ab) = log a + log b

then that means you can expand ln(1000x) as ln x + ln 1000 and since
ln 1000 is just a constant you will get just 1/x as the derivative, the same follows for ln(.001x)

7. Dec 16, 2005

### Robokapp

oh. Yea...forgot about that. you have a 1/ times dx/dx or times 1 and a 0 times dx/dx or 0*1+1*x^-1=dy/dx

I knew it's something really insignifficant...just didn't think it trough. :surprised

oh also about e wasn't e supposed to be calculated by the formula

(1+1/x)^x for large numbers? The problem with that is that it's never quite defined as a constant. If you plug in the graph of (1+1/x)^x for a range from 0 to 10^25 (Yea, I got a lot of free time to mess up with my TI83) you'd see it goes up, down, up again and then it keeps on growing until it gets out of range. Also if you take the derivative of (1+1/x)^x you'll lnotice that it indicates a behaviour far from constant...

plug in

y1=e
y2=(1+1/x)^x

for x-max = 100 and you'll see them be undistinctively overlapped.
go to x=10^20 and you'll see the y1 go up and down across the y2

Last edited: Dec 16, 2005
8. Dec 16, 2005

### d_leet

Are you attempting the product rule on that, dont even bother. suppose we have
ln ax

d/dx[ln u] = du/u
so if u = ax
the du = a
and u = ax
so we have ln ax = a/ax = 1/x.

As top the second part of your post e is defined as hypermorphism showed earlier, however e is also equal to
the limit as n goes to infinity of (1 + 1/n)^n which is precisely what you had. You can find that this limit is e using l'hospital's rule but it's a somewhat circular definition since the best way to find that limit is using the natural logarithm.