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Math Help - Finding the partial derivatives for ...

  1. #1
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    Question Finding the partial derivatives for ...

    f(x,y)= arctan(y/x)

    Im familiar with "outer" and "inner" derivates and usually know how to work them. I also know that the derivative of arctan = 1/(1+x^2)

    so i get

    f'x= 1/(1+x^2) * (y/x) * ??

    What is the x-derivative of y/x ?

    I also cannot seem to find the partial derivates to:

    f(x,y)= (x+y)/(x-y)

    Help would be highly appreciated...
    Last edited by tinyone; September 19th 2010 at 07:16 AM.
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  2. #2
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    When you calculate the partial derivative with respect to x, you treat y as a constant. Likewise when you differentiate with respect to y.

    So you get:
    \displaystyle f(x,y) = arctan \left( \frac{y}{x} \right) \Rightarrow f_y(x,y) = \frac{1}{1+ \left( \frac{y}{x} \right) ^2} \cdot \overbrace{\frac{\partial}{\partial y} \left( \frac{y}{x} \right)}^{ = \frac{1}{x}} \displaystyle = \frac{1}{x + \frac{y^2}{x}} \cdot \frac{x}{x} = \frac{x}{x^2 + y^2}

    You do the other one.
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  3. #3
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    To find \frac{\partial f}{\partial x}, take the derivative keeping y constant.

    So let u = \frac{y}{x} so that f = \arctan{u}.


    \frac{\partial u}{\partial x} = -\frac{y}{x^2}.


    \frac{df}{du} = \frac{1}{1 + u^2}

     = \frac{1}{1 + \left(\frac{y}{x}\right)^2}

     = \frac{1}{1 + \frac{y^2}{x^2}}

     = \frac{1}{\frac{x^2 + y^2}{x^2}}

     = \frac{x^2}{x^2 + y^2}.


    Therefore \frac{\partial f}{\partial x} = -\frac{y}{x^2}\left(\frac{x^2}{x^2 + y^2}\right)

     = -\frac{y}{x^2 + y^2}.


    To find \frac{\partial f}{\partial y}, take the derivative with respect to y keeping x fixed.

    Let u = \frac{y}{x} so that f = \arctan{u}.


    \frac{\partial u}{\partial y} = \frac{1}{x}.


    \frac{df}{du} = \frac{1}{1 + u^2}

     = \frac{1}{1 + \left(\frac{y}{x}\right)^2}

     = \frac{x^2}{x^2 + y^2}.


    Therefore \frac{\partial f}{\partial y} = \frac{1}{x}\left( \frac{x^2}{x^2 + y^2}\right)

     = \frac{x}{x^2 + y^2}.
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  4. #4
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    Quote Originally Posted by tinyone View Post
    f(x,y)= arctan(y/x)

    Im familiar with "outer" and "inner" derivates and usually know how to work them. I also know that the derivative of arctan = 1/(1+x^2)

    so i get

    f'x= 1/(1+x^2) * (y/x) * ??

    What is the x-derivative of y/x ?

    I also cannot seem to find the partial derivates to:

    f(x,y)= (x+y)/(x-y)

    Help would be highly appreciated...
    f(x, y) = \frac{x + y}{x - y}.


    \frac{\partial f}{\partial x} = \frac{(x - y)\frac{\partial}{\partial x}(x + y) - (x + y)\frac{\partial}{\partial x}(x - y)}{(x - y)^2}

     = \frac{(x - y) - (x + y)}{(x - y)^2}

     = -\frac{2y}{(x - y)^2}.


    \frac{\partial f}{\partial x} = \frac{(x - y)\frac{\partial}{\partial y}(x + y) - (x + y)\frac{\partial}{\partial y}(x - y)}{(x - y)^2}

     = \frac{(x - y) + (x + y)}{(x - y)^2}

     = \frac{2x}{(x - y)^2}.
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  5. #5
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    Thank you both so much, this forum is amazing.

    PS. If anybody else is eager to help Ive got another problem I can't solve

    Determine x*(f'x)+y*(f'y)+z*(f'z)

    if : f(x,y,z)= ln(x^3+y^3+^z^3 - 3xyz)

    The answer is simply 3. But when I derivate with regards to x, y and z and put it into x*(f'x)+y*(f'y)+z*(f'z) I get something far more complicated which I cannot simplify.
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