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    I'm trying to differentiate x=\frac{2y}{x^2-y} with respect to x. I know I could multiply by the denominator, then differentiate, but I've decided to use the quotient rule, and for some reason I can't get to the right answer.

    After simplifying, I get: \frac{dy}{dx}=\frac{(x^2-y)^2+4xy}{2x^2}

    Am I along the right lines, or have I done something completely wrong?
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    I really wouldn't advise the quotient rule with two variables on the bottom. I don't know what would happen but I can't imagine it's nice.
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    Well, it should work, should it not?
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    Could you expand your steps to see exactly what you did?

    I'm not entirely sure what you've done.
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    x=\dfrac{2y}{x^2-y}

    u=2y\Rightarrow\dfrac{du}{dx}=2\  dfrac{dy}{dx}

    v=x^2-y\Rightarrow\dfrac{dv}{dx}=2x-\dfrac{dy}{dx}

    1=...
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    1=\dfrac{(x^2-y)\frac{dy}{dx}-2y(2x-\frac{dy}{dx})}{(x^2-y)^2}=\dfrac{x^2\frac{dy}{dx}-y\frac{dy}{dx}-4xy+2y\frac{dy}{dx}}{(x^2-y)^2}

    Then simplified slightly further..
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    you forgot the 2 from 2y
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    As far as i can tell you can't differentiate with respect to x at the very start because when you differentiate the "x" on the left hand side you end up with dx/dx instead of dy/dx
    Just wondering, how did you put that formula up on your post?
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    Code:
    x=2y/(x2-y)
    
    x3-yx=2y
    
    x3-yx-2y=0
    
    differentiate --> 3x2-(dy/dx)(x)-y-2dy/dx=0
    
    3x2-y=dy/dx(x+2)
    
    dy/dx=(3x2-y)/(x+2)
    I'm new here so I'm not sure how to show it as a formula or anything, but that's what I got!
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    (Original post by Brecon)
    Code:
    x=2y/(x2-y)
    
    x3-yx=2y
    
    x3-yx-2y=0
    
    differentiate --> 3x2-(dy/dx)(x)-y-2dy/dx=0
    
    3x2-y=dy/dx(x+2)
    
    dy/dx=(3x2-y)/(x+2)
    I'm new here so I'm not sure how to show it as a formula or anything, but that's what I got!
    I got this also and comparing the values when x=1, y=0 from ours with the OP's I think his is wrong.
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    (Original post by Pheylan)
    x=\dfrac{2y}{x^2-y}

    u=2y\Rightarrow\dfrac{du}{dx}=2\  dfrac{dy}{dx}

    v=x^2-y\Rightarrow\dfrac{dv}{dx}=2x-\dfrac{dy}{dx}

    1=...
    Thanks .
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    No problem, hope that helped out a little :P This was one of my favourite topics last year.
 
 
 
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