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    Find and Classify the stationary values of 4xy^3=1+4x^2y^2
    I've found y' and the points of the stationary points to be (1/2,1) and (-1/2,-1).

    When finding the nature of the S.Pts I get both as minima, but the answer is a minimum at (1/2,1) and maximum at (-1/2,-1).
    I found y'' = 2y/(3y-4x), is this correct?


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    Local maximum at (1/2,1) and local minimum at (-1/2,-1). Workings?
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    (Original post by B_9710)
    Local maximum at (1/2,1) and local minimum at (-1/2,-1). Workings?
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    Don't try and find the second differential here, just test the gradient either side of the stationary point
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    (Original post by Bananapeeler)
    Don't try and find the second differential here, just test the gradient either side of the stationary point
    Doesn't that only work when you have the only x terms in the 1st derivative?


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    Couldn't you just pick an x value, plug it into your initial equation to solve for y and then put them both in your first derivative?
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    (Original post by Bananapeeler)
    Couldn't you just pick an x value, plug it into your initial equation to solve for y and then put them both in your first derivative?
    The equation is 4xy^3=1+4x^2y^2 so substituting a value for x would leave you with a cubic I believe. I think it'd be much easier to find the 2nd derivative in this case.


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    Sure, whatever works for you - it's just that solving a cubic seems easier than using the quotient rule implicitly, which has a lot more that could go wrong
 
 
 
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