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    How would you find for this question:

    Here's what I've done so far:



    Is this correct? If so, how do you make dy/dx the subject?

    Thanks
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    You can also do it by expanding those brackets
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    Which brackets do you mean?

    Is the working that I've done correct?
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    Looks right to me.

    But what do I know, I just started doing this :P
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    looks fine to me

    4-4\frac{dy}{dx}(x-y)^3=1+\frac{dy}{dx}

    \frac{dy}{dx}(-4(x-y)^3-1)=-3

    \frac{dy}{dx}=\frac{-3}{-4(x-y)^3-1}

    \frac{dy}{dx}=\frac{3}{4(x-y)^3+1}
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    (Original post by Pheylan)
    looks fine to me

    4-4\frac{dy}{dx}(x-y)^3=1+\frac{dy}{dx}

    \frac{dy}{dx}(-4(x-y)^3-1)=-3

    \frac{dy}{dx}=\frac{-3}{-4(x-y)^3-1}

    \frac{dy}{dx}=\frac{3}{4(x-y)^3+1}
    The answer at the back of the book is:

    \frac{dy}{dx}=\frac{4(x-y)^3-1}{1+4(x-y)^3}

    Is this wrong?
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    your working so far is correct, I think the next step is to multiply out or expand all the brackets, then get all the terms that are multiples of dy/dx on 1 side, the others on the other side, then you can factorise one side as they are all multiples of dy/dx. andonce you have factorised, make dy/dx the subject of the formula, by dividing ( i hope this helps im not very good at explanations)
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    (Original post by Pheylan)
    looks fine to me

    4-4\frac{dy}{dx}(x-y)^3=1+\frac{dy}{dx}

    \frac{dy}{dx}(-4(x-y)^3-1)=-3

    \frac{dy}{dx}=\frac{-3}{-4(x-y)^3-1}

    \frac{dy}{dx}=\frac{3}{4(x-y)^3+1}
    I don't think that's right.
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    (Original post by fireice7)
    The answer at the back of the book is:

    \frac{dy}{dx}=\frac{4(x-y)^3-1}{1+4(x-y)^3}

    Is this wrong?
    No, I got that answer, it's correct.

    It's just a case of being careful with your re-arranging.
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    (Original post by Skadoosh)
    No, I got that answer, it's correct.

    It's just a case of being careful with your re-arranging.
    How would you re-arrange it, (if pheylan's way is wrong) I'm still stuck at the line in my original post.
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    (Original post by Pheylan)
    looks fine to me

    4-4\frac{dy}{dx}(x-y)^3=1+\frac{dy}{dx}

    \frac{dy}{dx}(-4(x-y)^3-1)=-3

    \frac{dy}{dx}=\frac{-3}{-4(x-y)^3-1}

    \frac{dy}{dx}=\frac{3}{4(x-y)^3+1}
    fixed:

    (4-4\frac{dy}{dx})(x-y)^3=1+\frac{dy}{dx}

    4(x-y)^3-4\frac{dy}{dx}(x-y)^3=1+\frac{dy}{dx}

    \frac{dy}{dx}(-4(x-y)^3-1)=1-4(x-y)^3

    \frac{dy}{dx}=\frac{1-4(x-y)^3}{-4(x-y)^3-1}

    \frac{dy}{dx}=\frac{4(x-y)^3-1}{1+4(x-y)^3}
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    (Original post by Pheylan)
    fixed:

    (4-4\frac{dy}{dx})(x-y)^3=1+\frac{dy}{dx}

    4(x-y)^3-4\frac{dy}{dx}(x-y)^3=1+\frac{dy}{dx}

    \frac{dy}{dx}(-4(x-y)^3-1)=1-4(x-y)^3

    \frac{dy}{dx}=\frac{1-4(x-y)^3}{-4(x-y)^3-1}

    \frac{dy}{dx}=\frac{4(x-y)^3-1}{1+4(x-y)^3}
    how did you get from the first to the second line ??
 
 
 
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