about natural log

Calculus
The derivative of the natural logarithm function is:


圖片參考：http://upload.wikimedia.org/math/f/6/5/f65b3e8b30827a42efd0bf16f8a196c2.png

and by applying the change-of-base rule, the derivative for other bases is:


圖片參考：http://upload.wikimedia.org/math/9/f/3/9f3c73b3c4e8366aa669c8711a466e7e.png

The antiderivative of the logarithm is


圖片參考：http://upload.wikimedia.org/math/f/9/b/f9bc57d9d60beda31387aa5ab0edecf0.png



[edit] The natural logarithm in integration
The natural logarithm allows simple integration of functions of the form g(x) = f '(x)/f(x): an antiderivative of g(x) is given by ln(|f(x)|). This is the case because of the chain rule and the following fact:


圖片參考：http://upload.wikimedia.org/math/1/7/5/17548a0e9a92e13b63564cdef1e39669.png

In other words,


圖片參考：http://upload.wikimedia.org/math/5/d/2/5d278c710aa6319d38d1ce6eb717594a.png

and


圖片參考：http://upload.wikimedia.org/math/5/a/c/5ac11c895245be47203cafaec57f1f8d.png

Here is an example in the case of g(x) = tan(x):


圖片參考：http://upload.wikimedia.org/math/3/d/8/3d8507d6d9a91335dd8d37a2005f915e.png


圖片參考：http://upload.wikimedia.org/math/e/8/f/e8fa0130c9d536efa79655f02a492214.png

Letting f(x) = cos(x) and f'(x)= - sin(x):


圖片參考：http://upload.wikimedia.org/math/b/1/b/b1b0e05bca2ed3b0fe85eab2d5e117fb.png


圖片參考：http://upload.wikimedia.org/math/f/a/b/fab3ff5a35605d0e31511a4e66ff1125.png

where C is an arbitrary constant of integration.
The natural logarithm can be integrated using integration by parts:


圖片參考：http://upload.wikimedia.org/math/b/8/1/b8102a2b9538d0294db6bd878afd0f97.png

The definition of e is
e = lim h-?INF [(1+1/h)^h]
using a calculator, e^1 = 2.718... something like that.

(P.S. I use -? instead of an arrow in the forum.)

natural log is taking the log of base e.
i.e. ln y = log e (y) (the letter e is very small and at the bottom, I cannot put it there in the post)

From the definition above, we can prove the following (all proofs are done by myself without looking at other references):

Given y = f(x) = e^x, what is dy/dx=?
From the 1st principle,
dy/dx
= lim h-?0 [f(x+h)-f(x)]/h
= lim h-?0 [e^(x+h)-e^x]/h
= lim h-?0 e^x [e^h-1]/h
= e^x lim h-?0 [e^h-1]/h
= e^x lim h-?0 [e^h-1]/h (Let k = 1/h)
= e^x lim k-?INF k[e^(1/k)-1]
= e^x lim k-?INF k{[(1+1/k)^k]^(1/k)-1} (see the definition above)
= e^x lim k-?INF k{(1+1/k)^[k*(1/k)]-1} (law of indices)
= e^x lim k-?INF k{(1+1/k)-1}
= e^x lim k-?INF k(1/k)
= e^x lim k-?INF 1
= e^x.
Therefore, y = e^x, dy/dx = e^x itself.
Then, Int e^x dx = e^x + C.

Int -- integral sign (for the forum use)

If y = ln x, what is dy/dx?
y = ln x
e^y = x
Differentiate both sides with respect to x,
e^y (dy/dx) = 1 (chain rule and see the rule above)
x dy/dx = 1
dy/dx = 1/x
Therefore, y = ln x, dy/dx = 1/x
Then, Int 1/x dx = ln x + C as x GT 0
What happens if x is negative?
If x LT 0, Let t = -x, x = -t, dx/dt = -1,
Int 1/x dx = Int 1/(-t) (-1) dt
= Int 1/t dt
= ln t + C
= ln (-x) + C (Do not worry the negative sign inside the ln, since x LT 0)
In general, Int 1/x dx = ln abs(x) + C as x != 0
abs = absolute value
Combined the rules,
Int x^n dx = ln abs(x) + C if n = -1, otherwise, [x^(n+1)]/(n+1) + C

These are the most general basic formulas from natural log for calculus.
You can use these for doing other integrations:
e.g.
Let u = cos x
du/dx = - sin x
Doing Int tan x dx by using a substitution and natural log definitions:
Int tan x dx
= Int sin x/cos x dx
= Int -1/cos x (-sin x) dx
= Int -1/u du
= - ln abs(u) + C (see the definition above)
= - ln abs(cos x) + C

P.S.
GT -- greater than
LT -- less than
!= -- is not equal to