m1 derivaives f4

y=(8x^(¾)-3)(6-x^(¼))
y'
= [8x^(¾)-3][-¼x^(¼-1)] + [8(¾)x^(¾-1)][6-x^(¼)] (积法则)
= [8x^(¾)-3][-¼x^(-¾)] + [6x^(-¼)][6-x^(¼)]
= [8x^(¾)][-¼x^(-¾)] -3[-¼x^(-¾)]+ [6x^(-¼)][6] -x^(¼)[6x^(-¼)]
= -2 +¾x^(-¾)+ 36x^(-¼)-6
=¾x^(-¾)+ 36x^(-¼)-8

http://www.wolframalpha.com/input/?i=derivative+y%3D%288x%5E%28%C2%BE%29-3%29%286-x%5E%28%C2%BC%29%29


y=(x^3-2x)/(2x-1)
y'
=[(3x-2)(2x-1)-(x³-2x)(2)]/(2x-1)² (商法则)
=[(6x²-7x+2)-(2x³-4x)]/(2x-1)²
=[6x²-3x+2-2x³]/(2x-1)²
.
.←中間我都唔系好知点做
.
=(4x³-3x²+2)/(1-2x)²

http://www.wolframalpha.com/input/?i=derivative+y%3D%28x%5E3-2x%29%2F%282x-1%29+


以上我系自己做,做完睇wolframalpha,而基本原理唔做啦!唔明,烦请來信,thx.

2013-12-05 18:52:30 補充：
回答者：魂( 小學級 5 級 ) 做少咗D

eg.
y=xⁿ

y'=nx^(n-1)

Find the derivative of each of the following functions with respect to x.
y=(8x^3/4-3)(6-x^1/4)

Assume your question is [8x^(3/4)-3][6-x^(1/4)], the derivative y':
y' = [8x^(3/4)-3][-1x^(1/4-1)] + [8x^(3/4-1)][6-x^(1/4)] (product rule)
= -[8x^(3/4)-3][x^(-3/4)] + [8x^(-1/4)][6-x^(1/4)]
//= -[8x^(3/4)-3]/[x^(3/4)] + [6-x^(1/4)]/[8x^(1/4)]//

y=(x^3-2x)/(2x-1)
The derivative y':
y' = {[3(x^2)-2(1)](2x-1) - (x^3-2x)(2*1)]/[(2x-1)^2]
= [(3x^2-2)(2x-1) - 2(x^3-2x)]/[(2x-1)^2]
= (6x^3-3x^2-4x+2-2x^3+4x)/[(2x-1)^2]
//= (4x^3-3x^2+2)/(4x^2-4x+1)//