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Math Help - Implicit Partial Deriviative

  1. #1
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    Implicit Partial Deriviative

    Stuck on this one also

    Given that y=y(x,z) is defined implicitly by the relation

     y^3 + z\sqrt{x} - ln(xz^{2}-y) = \frac{16}{x^2} +24

    compute  \frac{dy}{dz} at the point (4,-3,1)

    Thanks!
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  2. #2
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    Quote Originally Posted by Gotovina7 View Post
    Stuck on this one also

    Given that y=y(x,z) is defined implicitly by the relation

     y^3 + z\sqrt{x} - ln(xz^{2}-y) = \frac{16}{x^2} +24

    compute  \frac{dy}{dz} at the point (4,-3,1)

    Thanks!
    Treat x as a constant and differentiate implicitly as you would in the single variable case. Half the answer: the derivative of the right hand side is zero.
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    Quote Originally Posted by mr fantastic View Post
    Treat x as a constant and differentiate implicitly as you would in the single variable case. Half the answer: the derivative of the right hand side is zero.
    Ok, should I split up that ln?
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    Quote Originally Posted by Gotovina7 View Post
    Ok, should I split up that ln?
    xz^2 - y cannot be factorised. Since you're not logging a pure product, you can't split the log up ....
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  5. #5
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    Quote Originally Posted by Gotovina7 View Post
    Stuck on this one also

    Given that y=y(x,z) is defined implicitly by the relation

     y^3 + z\sqrt{x} - ln(xz^{2}-y) = \frac{16}{x^2} +24

    compute  \frac{dy}{dz} at the point (4,-3,1)

    Thanks!
    the code for a partial is \partial y

    3y^2\frac{\partial y}{\partial z}+\sqrt{x}-\frac{2xz-\frac{\partial y}{\partial z}}{xz^2-y}=0

    eval @ (4,-3,1)

    3(-3)^2\frac{\partial y}{\partial z}+\sqrt{4}-\frac{2(4)(1)-\frac{\partial y}{\partial z}}{(4)(1)^2-(-3)}=0

    27 \frac{\partial y}{\partial z} +2 -\frac{8-\frac{\partial y}{\partial z}}{7}=0

    189 \frac{\partial y}{\partial z} +14 -8 +\frac{\partial y}{\partial z}=0 \iff 190 \frac{\partial y}{\partial z} =-6 \iff \frac{\partial y}{\partial z}=\frac{-3}{95}
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