differentiation of 'e'

發問者 is more interested in the use of formula rather than derivation of the formula. Well, I may be wrong

y = b^x

dy
--- = b^x. ln b [b is a numerical constant and ln b is natural logarithm of b ]
dx

On the calculator, there are log (to the base of 10) and ln (to the base of e , e = 2.71828..)
e.g. y = 2^x,
dy/dx = 2^x ln 2 = 2^x (0.6931) = 0.6931 2^x

y = 3^x,
dy/dx = 3^x ln 3 = 2^x (1.0986) = 1.0986 3^x

It seems reasonable that there is a number, e, for which ln e =1, and that number lies between 2 and 3. That number is e = 2.71828…. It is called the natural number
e is the number such that lim (e^h – 1)/h = 1 as h approaches 0.
e is actually defined as the lim (1 + u)^(1/u) as u approaches zero.
It is given by: e = 1+ 1+ 1/(2!) + 1/(3!) + 1/(4!) + …..

Just remember this formula, e is raised to the power x.
y = e^x

dy
--- = e^x.
dx

Another formula you have to remember
y = e^g(x) where g(x) is a function of x
dy/dx = e^g(x) g’(x)

Example:
y = e^(x^2 + 5x)
g(x) = x^2 + 5x, so g’(x) = 2x + 5
dy/dx = e^g(x) g’(x)
dy/dx = e^( x^2 + 5x) (2x + 5)
dy/dx = (2x + 5) e^(x^2 + 5x)

To回答者：Henry1989
意見者：Ho-yin. is right. I couldn’t agree more.
Your derivations in both methods are basically flawed.
You have to prove lim (e^h – 1)/h = 1 as h approaches zero ~ not by L' Hospital Rule

Henry1989的prove 都不成立

1. 用 l'hopita rule在證明的過程中使用了要證明的命題.

2. 這個證明看來沒有問題, 不過它用到 ln(x) 的微分.
如果要證明ln(x) 的微分的公式... 恐怕又是先要用到 e^(x) 的微分.

e=2.7.......
係constant
所以d(e^2),d(e^3)之類全部都只係當d(constant)咁計,即0

我想問係唔係係即係power要有個x字出現先用得??
答: 係.


How to prove it?

Method 1: L' Hospital Rule

d(e^x)/dx
= lim h->0 [e^(x+h) - e^x]/h
= lim h->0 e^x (e^h -1)/h
= lim h->0 e^x * lim h->0 (e^h -1)/h
= e^x * lim h->0 e^h / 1
= e^x * e^0
= e^x * 1
= e^x

Method 2

Let y = e^x
ln y = ln (e^x)
= x lne
= x * 1
ln y = x
d(ln y)/dx = dx/dx
1/y dy/dx = 1
dy/dx = y
= e^x

So, d(e^x)/dx = e^x

2011-06-18 12:29:56 補充：
唔好意思呀, 我用錯左方法.

至於點樣去証, 之前有人解答過:
http://hk.knowledge.yahoo.com/question/question?qid=7007012103367