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Implicit differentiation (algebra question)

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    I'm differentiating this function:

     (x-y)^4 = x + y + 5

    So far I've got this:

     4(x-y)^3 = \dfrac{1+y'}{1-y'}

    I need to get an expression for y' but I'm really struggling to separate them out. Could someone show me how to do this. I'm sure it's simple.

    Thanks!
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    Have you tried cross multiplying? It should be possible to collect like terms for dy/dx and then use subject formula.
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    If I asked you to solve this equation for a, could you do it?

    \displaystyle 5 = \frac{1+a}{1-a}

    Your question is no different.
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    (Original post by aurao2003)
    Have you tried cross multiplying? It should be possible to collect like terms for dy/dx and then use subject formula.
    Thanks for the reply. Do you mean multiplying out the  4(x-y)^3(1-y') and then collecting the y' terms? I thought about doing that, but is there another way that would be less time consuming in the exam?
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    (Original post by Swayum)
    If I asked you to solve this equation for a, could you do it?

    \displaystyle 5 = \frac{1+a}{1-a}

    Your question is no different.
    Thanks. So are you saying that multiplying out the brakets is the only way in this case?
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    (Original post by JeremyB)
    Thanks for the reply. Do you mean multiplying out the  4(x-y)^3(1-y') and then collecting the y' terms? I thought about doing that, but is there another way that would be less time consuming in the exam?
    Well, you didn't need to divide through by 1-y' in the first place, right? So that saves you a couple of lines of working out, but no there's nothing shorter than rearranging the equation for y'.
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    (Original post by JeremyB)
    Thanks for the reply. Do you mean multiplying out the  4(x-y)^3(1-y') and then collecting the y' terms? I thought about doing that, but is there another way that would be less time consuming in the exam?
    Also, just to clarify, multiplying out the above just means getting:

     4(x-y)^3(1-y') = 4(x-y)^3 - 4y'(x-y)^3

    You obviously wouldn't expand out (x-y)^3.
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    (Original post by Swayum)
    Also, just to clarify, multiplying out the above just means getting:

     4(x-y)^3(1-y') = 4(x-y)^3 - 4y'(x-y)^3

    You obviously wouldn't expand out (x-y)^3.
    Ah, thanks for that. I was thinking I had to multiply out the (x-y)^3.

    So would this give me:

    y' = \dfrac{1-4(x-y)^3}{4(x-y)^3 + 1} ?
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    (Original post by JeremyB)
    Ah, thanks for that. I was thinking I had to multiply out the (x-y)^3.

    So would this give me:

    y' = \dfrac{1-4(x-y)^3}{4(x-y)^3 + 1} ?
    Nope, the numerator should be 4(x-y)^3 - 1 I think (although I could be wrong).
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    (Original post by Swayum)
    Nope, the numerator should be 4(x-y)^3 - 1 I think (although I could be wrong).
    Thanks a lot for your help.

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