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    (Original post by student92)
    Hey, i got another question... if u could help me: find dy/dx of 3/x-2 - 2/x+3 when x is -1. The answer is 1/6 but i keep getting -1/6
    1/6 is correct.

    Try rewriting 3/x-2 as 3(x-2)^-1 and the same with the other term. See if you can go from there.
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    (Original post by Hippysnake)
    1/6 is correct.

    Try rewriting 3/x-2 as 3(x-2)^1 and the same with the other term. See if you can go from there.
    BRACKETS PLEASE
    3/(x-2) = 3(x-2)^{-1}
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    (Original post by JamesyB)
    Have you rewitten it as 3(x - 2)^{-1} - 2(x + 3)^{-1} and used the chain rule?
    yes
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    (Original post by student92)
    yes
    Have you checked to make sure all your signs are correct?
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    (Original post by Hippysnake)
    1/6 is correct.

    Try rewriting 3/x-2 as 3(x-2)^-1 and the same with the other term. See if you can go from there.
    i have rewritten it like that... after differentiating i get: 3(x-2 )^-2 - 2(x+3)^-2
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    (Original post by student92)
    i have rewritten it like that... after differentiating i get: 3(x-2 )^-2 - 2(x+3)^-2
    Sure it's '-2' and not '+2'?

    if you differentiate

    -2(x+3)^-1 you get

    -1*-2(x+3)^-1
    which is +2(x+3)^-2
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    (Original post by Hippysnake)
    Sure it's '-2' and not '+2'?

    if you differentiate

    -2(x+3)^-1 you get

    -1*-2(x+3)^-1
    which is +2(x+3)^-1
    This is not correct. The answer is

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    (Original post by steve2005)
    This is not correct. The answer is

    I think I'll just keep to myself....I got the right answer on paper, just forgot to put the -2 term up.

    Ah well, OP probably figured out how to do it on her own anyway.
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    (Original post by student92)
    Hey, how would you work this out: The integral of 4 / 2x - 3 between the limits x is 6 and 2?
    \displaystyle{\int_2^{6}{\dfrac{  4}{2x+3}} = 4\int{\dfrac{1}{2x-3}} = 4\bigg[\dfrac{1}{2}\log\left(2x-3\right)\bigg]_2^{6}}

= 4\left[\dfrac{1}{2}\log\left\{2(6)-3\right\}-\dfrac{1}{2}\log\left\{(2)(2)-3)\right\}\right]  = 4\left[\dfrac{1}{2}\log(9)-\dfrac{1}{2}\log(1)\right] = 4\left[\dfrac{1}{2}\log(9)\right]

    = 4\log\sqrt{9} = 4\log(3) = \log(3)^4 = \log(81).

    I've an exam on this few days to go. I hope you guys don't mind me solving an already solved question for practice. Haha.
 
 
 
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