一條Derivative問題(急需2小時內解決!)

a. [√2x-1 - √2(x+△x)-1 ][√2x-1 + √2(x+△x)-1 ]

= (√2x-1)^2 - (√2(x+△x)-1)^2 (By (a - b)(a + b) = a^2 - b^2)

= (2x - 1) - [2(x+△x) - 1]

= 2x - 1 - 2x - 2△x + 1

= -2△x


b. f(x) = 1 / √(2x-1)

f'(x)

= lim △x→ 0 [f(x + △x) - f(x)] / △x

= lim △x→ 0 [1/√(2(x + △x)-1) - 1/√(2x-1)] / △x

= lim △x→ 0 [√2x-1 - √2(x+△x)-1 ] / △x√(2x-1)√(2(x + △x)-1)

= lim △x→ 0 [√2x-1 - √2(x+△x)-1 ][√2x-1 + √2(x+△x)-1 ] / △x√(2x-1)√(2(x + △x)-1)[√2x-1 + √2(x+△x)-1 ]

= lim △x → 0 -2△x / △x√(2x-1)√(2(x + △x)-1)[√2x-1 + √2(x+△x)-1 ]

= -2 lim △x → 0 1 / √(2x-1)√(2(x + △x)-1)[√2x-1 + √2(x+△x)-1 ]

= -2 / √(2x-1)√(2(x + 0)-1)[√2x-1 + √2(x+0)-1 ]

= -2 / (2x - 1)(2√2x-1)

= -1 / (2x - 1)^(3/2)

2009-02-19 20:35:03 補充：
If you cannot see the above expressions clearly, please go to the following link
http://i718.photobucket.com/albums/ww188/physicsworld2009/physicsworld01Feb192033.jpg

I think this one is much clearer to comprehend.