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Math Help - Partial Derivatives

  1. #1
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    Partial Derivatives

    f(x,y)=x^2y + xsiny -ln[x-y^2]

    Find fy and fxy

    I found fy but to get xy you take the x derivative, which I found to be 2xy+siny-1/(x-y^2) and then take the y. The answer is 2x-xsiny-2y/(x-y^2)^2

    This is a practice midterm problem and I messed up somewhere because I can't get xsiny from siny. Could someone walk me through this or point me to where I went wrong in calculating fxy?
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  2. #2
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    Quote Originally Posted by sfgiants13 View Post
    f(x,y)=x^2y + xsiny -ln[x-y^2]

    Find fy and fxy

    I found fy but to get xy you take the x derivative, which I found to be 2xy+siny-1/(x-y^2) and then take the y. The answer is 2x-xsiny-2y/(x-y^2)^2

    This is a practice midterm problem and I messed up somewhere because I can't get xsiny from siny. Could someone walk me through this or point me to where I went wrong in calculating fxy?
    All I can say is that the "answer", 2x- xsiny- 2y/(x- y^2)^2 is wrong.

    You can do this either of two ways. Since you were asked to find f_y and had already found that f_y= x^2+ x cos(y)+\frac{2y}{x-y^2} just differentiate that with respect to x: f_{yx}= 2x+ cos(y)- \frac{2y}{(x- y^2)^2}.

    Or, first find f_x= 2xy+ siny- \frac{1}{x-y^2} and differentiate that with respect to y: f_{xy}= 2x+ cos y- \frac{2y}{(x- y^2)^2}.

    (As long as the second derivatives are continuous, the "mixed derivatives", f_{xy} and f_{yx}, are equal.)
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    All I can say is that the "answer", 2x- xsiny- 2y/(x- y^2)^2 is wrong.

    You can do this either of two ways. Since you were asked to find f_y and had already found that f_y= x^2+ x cos(y)+\frac{2y}{x-y^2} just differentiate that with respect to x: f_{yx}= 2x+ cos(y)- \frac{2y}{(x- y^2)^2}.

    Or, first find f_x= 2xy+ siny- \frac{1}{x-y^2} and differentiate that with respect to y: f_{xy}= 2x+ cos y- \frac{2y}{(x- y^2)^2}.

    (As long as the second derivatives are continuous, the "mixed derivatives", f_{xy} and f_{yx}, are equal.)
    That's what I got actually but the solutions sheet gave the wrong answer apparently. Thanks!
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