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
微積分問題 (5分)
2011-09-27 9:50 pm
回答 (2)
2011-10-01 6:38 pm
✔ 最佳答案
[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
2011-09-27 10:21 pm
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
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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