中四Amaths 從基本原理求導數

AS follow~~~~~~~~~~~~~~~~~~~~
AS follow~~~~~~~~~~~~~~~~~~~~

圖片參考：http://i302.photobucket.com/albums/nn113/ken123572003/5567ce17.jpg

f(x) =1/sqrt(3x -1)
f(x + h) = 1/sqrt[3(x +h) -1].
f( x + h) -f(x) = 1/sqrt(3x +3h -1) -1/sqrt(3x -1)
= [sqrt(3x -1) - sqrt(3x +3h -1)]/sqrt[(3x +3h -1)(3x -1)]
= [sqrt(3x -1) - sqrt(3x +3h -1)][sqrt(3x -1) + sqrt(3x +3h -1)]/[sqrt(3x +3h -1)(3x -1)][sqrt(3x -1) +sqrt(3x +3h -1)]
= [(3x -1) - (3x +3h -1)]/[sqrt(3x +3h -1)(3x -1)][sqrt(3x -1) + sqrt(3x +3h -1)]
=-3h/[sqrt(3x +3h -1)(3x -1)][sqrt(3x -1) + sqrt(3x +3h -1)]
Therefore, [f(x +h) -f(x)]/h = -3/[sqrt(3x +3h -1)(3x -1)][sqrt(3x -1) + sqrt(3x+3h-1)]......................(1)
Lim h tends to 0, that is putting h in (1) to 0, we get -3/[sqrt(3x -1)^2][2sqrt(3x-1)]
That is dy/dx = -3/2(3x -1)sqrt(3x -1) = -3/[2(3x -1)^3/2].

基本原理係咪果個delta x tend to zero 果個?
宜家設delta x 做h(因為易睇d)
設f(x)=1/(3x-1)

lim h－＞0 [f(x h)-f(x)]/h
=lim h－＞0 [1/(3x 3h-1)-1/(3x-1)]/h (將上面果舊通分母,做加減)
=lim h－＞0 [-3h/(3x 3h-1)(3x-1)]/h
=lim h－＞0 -3/(3x 3h-1)(3x-1)
=-3/(3x-1)^2