微積分 differentiate e^x

In order to prove this by first principle, we need the following expansion

e^x =1 +x +(x^2 /2!) +(x^3 /3!)+(x^4 /4!)+.....+(x^n /n!) +.......

Now,
d/dx e^x
=lim_(Δx to 0){ e^(x+Δx) - e^x}/Δx
=lim_(Δx to 0){ e^x (e^Δx) - 1)}/Δx
=lim_(Δx to 0){ e^x (Δx +(Δx^2 /2!) +.......)}/Δx
=lim_(Δx to 0){ e^x (1 +(Δx /2!) +.......)}
= e^x (1 +(0 /2!) +.......)
= e^x (1 +0 )
=e^x


圖片參考：http://us.a2.yahoofs.com/users/45b36d09zd8f9dafd/1bf2scd/__sr_/ffefscd.jpg?phwm3sFBPgKHle.3


2007-01-22 22:04:55 補充：
d/dx e^x=lim_(Δx -> 0){ e^(x+Δx) - e^x}/Δx=lim_(Δx -> 0){ e^x (e^Δx - 1)}/Δx=lim_(Δx -> 0){ e^x ( 1 +Δx +(Δx^2 /2!) +....... -1)}/Δx

2007-01-22 22:05:22 補充：
=lim_(Δx -> 0){ e^x (Δx +(Δx^2 /2!) +.......)}/Δx=lim_(Δx -> 0){ e^x (1 +(Δx /2!) +.......)}= e^x (1 +(0 /2!) +.......)= e^x (1 +0 )=e^x

e^x = 1 + x + x^2/(1*2) + x^3/(1*2*3) + ..... + x^n/n!
de^x = d (1 + x + x^2/(1*2) + x^3/(1*2*3) + ..... + x^n/n!)
= 0 + 1 + x + x^2/(1*2) + x^3/(1*2*3) + ..... + x^n/n!
= e^x

Prove that d/dx ex = ex

Let y = ex

ln y = ln (ex)

ln y = x (ln e) 【Using identity ln(ab) = b(ln a)】

ln y = x (1) 【Using the fact ln e = 1】

ln y = x

Differentiating both sides

d(ln y)/dx = dx/dx

(1/y)(dy/dx) = 1

dy/dx = y

dy/dx = ex

So d(ex)/dx = ex

2007-01-21 17:42:33 補充：
小小補充：d(e^x)/dx = e^x 是一個已知的微分 (differentiation) 恆等式。平時運算時都可以使用，證明這條恆等式比較少見喔！ ^^