locmsimon asked in 科學及數學數學 · 9 years ago

微積分問題 (5分)

Let y=f(x) be a differentiavble function, where f(0)=1, f(1)=3, f ' (0)=2 and f ' (1)=-1.

(a) Find the value of d/dx((2f(x)/(x+1)) when x=0.

(b) Find the value of d/dx{ f [ (1-x)/(1+x) ] } when x=1.

answer (a)=2

(b)=12

請解釋

thanks

Update:

sorry,

b答案係 -1

Update 2:

可唔可以詳細講述個步驟?

點解第一題用product rule,同埋步驟兼解釋?

第2題都係,thanks

2 Answers

Rating
  • ?
    Lv 7
    9 years ago
    Favorite Answer

    [2f(x)/(x + 1)]'= [2f'(x)(x + 1) - 2f(x)]/(x + 1)^2Sub. x = 0The value of d/dx((2f(x)/(x+1)) is 2(b) Let y = (1 - x)/(1 + x)y' = [-(1 + x) - (1 - x)]/(1 + x)^2= -2/(1 + x)^2Since {f[(1 - x)/(1 + x)]}'= f'(y) * y'Sub. x = 1, y = 0 => d/dx{ f [ (1-x)/(1+x) ] } = 2 * (-1/2)= -1

  • 9 years ago

    a)d/dx((2f(x)/(x+1)) =2f'(x)/(x+1) + (-2f(x)/(x+1)^2) <-- product rule

    when x=0, 2f'(0)-2f(0)=2(2)-2(1)=2

    b) d/dx{ f [ (1-x)/(1+x) ] } = f'[ (1-x)/(1+x) ] d/dx[ (1-x)/(1+x) ] <-- chain rule

    =f'[ (1-x)/(1+x) ][-(1-x)/(1+x)^2-1/(1+x)] <-- product rule

    when x=1, f'(0)(-1/2)=(2)(-1/2)=-1

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