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# Differentiation watch

1. I've been killing myself trying to do this one:-

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

I've tried the quotient rule :-

dy/dx = [((x-1)^(1/2)).2 - (1/2(x-1)^(-1/2)).2x] / (x-1)

upon cancelling down i get [x(x-1) - 2(x-1)(x-1)] / [(x-1)^(3/2)]

the books answer is (x-2) / (x-1)^(3/2)

Can anyone help?
2. howdy

after differentiating, factorise in terms of (x-1)^-1/2
then you get that * by 2(x-1)-x on the top
and if you take the factor to the bottom, multiply through you get the x-1)^3/2 at the bottom and simplifiying the top gives x-2

3. (Original post by robfinlay)
I've been killing myself trying to do this one:-

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

I've tried the quotient rule :-

dy/dx = [((x-1)^(1/2)).2 - (1/2(x-1)^(-1/2)).2x] / (x-1)

upon cancelling down i get [x(x-1) - 2(x-1)(x-1)] / [(x-1)^(3/2)]

the books answer is (x-2) / (x-1)^(3/2)

Can anyone help?
This is right. now split it up into two separate terms to get:

[2(x-1)^.5]/(x-1) - x/[(x-1)^1.5]

Multiply LH term by [(x-1)^.5]/[(x-1)^.5] to get them both over the denominator (x-1)^1.5. Then it should all work out.
4. hmm, still dont understand. my arithmetic is messy
5. ahhh thanks alot.

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Updated: February 4, 2006
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