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Math Help - need help on partial derivatives

  1. #1
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    need help on partial derivatives

    1.)f(x,y) = arctan(y/x) fx(2,3)



    how would I do this?



    2.) f(x,y) = ln(x+ sqrt(x^2+y^2) ) fx(3,4)
    1/(+ sqrt(x^2+y^2) * 1+ 1/2(x^2 + y^2) * 2x
    is that correct for #2?



    Thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by khuezy View Post
    f(x,y) = arctan(y/x) fx(2,3)

    how would I do this?
    Thanks
    First, partially differentiate f(x,y) with respect to x, and then evaluate it at (2,3):

    f(x,y)=\tan^{-1}\left(\frac{y}{x}\right)\implies\frac{\partial f}{\partial x}=\frac{1}{1+\left(\frac{y}{x}\right)^2}\cdot \left(-\frac{y}{x^2}\right)=-\frac{y}{x^2+y^2}

    Now what is f_x\left(2,3\right)??

    --Chris
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  3. #3
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    thanks

    i just plug in x and y into that equation right?
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  4. #4
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    thanks

    how would I do this?



    2.) f(x,y) = ln(x+ sqrt(x^2+y^2) ) fx(3,4)
    1/(+ sqrt(x^2+y^2) * 1+ 1/2(x^2 + y^2) * 2x
    is that correct for #2?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by khuezy View Post
    how would I do this?



    2.) f(x,y) = ln(x+ sqrt(x^2+y^2) ) fx(3,4)
    1/(x+ sqrt(x^2+y^2)) * 1+ (1/2(x^2 + y^2) * 2x)
    is that correct for #2?
    Note the slight modifications I made.

    It should be f_x(x,y)=\frac{1}{x+\sqrt{x^2+y^2}}\cdot\left(1+\f  rac{x}{\sqrt{x^2+y^2}}\right)

    Then plug in 3 for x and 4 for y.

    --Chris
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    First, partially differentiate f(x,y) with respect to x, and then evaluate it at (2,3):

    f(x,y)=\tan^{-1}\left(\frac{y}{x}\right)\implies\frac{\partial f}{\partial x}=\frac{1}{1+\left(\frac{y}{x}\right)^2}\cdot \left(-\frac{y}{x^2}\right)=-\frac{y}{x^2+y^2}

    Now what is f_x\left(2,3\right)??

    --Chris
    Quote Originally Posted by khuezy View Post
    i just plug in x and y into that equation right?
    Yes! What did you get for the answer?

    --Chris
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  7. #7
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    hi again

    Find all the second partial derivatives of
    v = xy/(x-y)

    could you quickly show me the steps to see if I did it correctly?
    Thanks again
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by khuezy View Post
    Find all the second partial derivatives of
    v = xy/(x-y)

    could you quickly show me the steps to see if I did it correctly?
    Thanks again
    First you need to find \frac{\partial f}{\partial x} and \frac{\partial f}{\partial y}

    f(x,y)=\frac{xy}{x-y}

    \frac{\partial f}{\partial x}=\frac{(x-y)y-xy(1)}{(x-y)^2}=-\frac{y^2}{(x-y)^2}

    \frac{\partial f}{\partial y}=\frac{(x-y)x-xy(-1)}{(x-y)^2}=\frac{x^2}{(x-y)^2}

    Now, there are four different second partial derivatives:

    \frac{\partial^2 f}{\partial x^2}, \frac{\partial^2 f}{\partial y^2}, \frac{\partial^2 f}{\partial y\partial x} and \frac{\partial^2 f}{\partial x\partial y}

    \frac{\partial^2 f}{\partial x^2}=-\frac{(x-y)^2(0)-y^2(2(x-y))}{(x-y)^4}=\frac{2y^2}{(x-y)^3}

    \frac{\partial^2 f}{\partial y^2}=\frac{(x-y)^2(0)-x^2(-2(x-y))}{(x-y)^4}=\frac{2x^2}{(x-y)^3}

    \frac{\partial^2 f}{\partial y\partial x}=-\frac{(x-y)^2(2y)-y^2(-2(x-y))}{(x-y)^4}=-\frac{2y(x-y)+2y^2}{(x-y)^3}=-\frac{2xy}{(x-y)^3}

    \frac{\partial^2 f}{\partial x\partial y}=\frac{(x-y)^2(2x)-x^2(2(x-y))}{(x-y)^4}=\frac{2x(x-y)-2x^2}{(x-y)^3}=-\frac{2xy}{(x-y)^3}

    Note, you should always see that \frac{\partial^2f}{\partial x\partial y}\equiv \frac{\partial^2f}{\partial y\partial x}

    Does this make sense?

    --Chris
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  9. #9
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    i got the fxx and fyy

    but I don't get how you did the fyy/fxx

    aren't you suppose to the the derivative of fx
    it looks like you took it from the f(x,y) instead of fx(x,y)
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by khuezy View Post
    but I don't get how you did the fyy/fxx

    aren't you suppose to the the derivative of fx
    it looks like you took it from the f(x,y) instead of fx(x,y)
    \frac{\partial{f(x,y)}}{\partial{y}^2}=\frac{\part  ial}{\partial{y}}\left\{\frac{\partial{f(x,y)}}{\p  artial{y}}\right\}

    Is that what you are asking?
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