about natural log

Question

i want to know how natural log can be used in differentiation and integration. if there is any formula, please tell me the prove, or give me some web site about this question.

i am only in F.5 level so i hope you can try to answer my question in a simpler way.

Answer 1

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

Answer 2

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