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Math Help - Partial Derivatives

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    Partial Derivatives

    Given the following function, find f_(xx), f_(xy), f_(yyx) [Note: f_(xx), for instance, is f subscript xx].

    f(x,y) = e^(4x) - sin(y^2) - sqrt(xy)

    Answers in book:

    f_(xx) = 16e^(4x) + (1/4)*sqrt(y/(x^3))
    f_(xy) = -1/(4*sqrt(xy))
    f_(yyx) = 1/(8*sqrt(x^3*y^3))
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    Quote Originally Posted by Ideasman View Post
    Given the following function, find f_(xx), f_(xy), f_(yyx) [Note: f_(xx), for instance, is f subscript xx].

    f(x,y) = e^(4x) - sin(y^2) - sqrt(xy)

    Answers in book:

    f_(xx) = 16e^(4x) + (1/4)*sqrt(y/(x^3))
    f_(xy) = -1/(4*sqrt(xy))
    f_(yyx) = 1/(8*sqrt(x^3*y^3))
    I'll do f_(xx), you can do the rest.

    To get a partial derivative with respect to x, treat all other variables as constants. So given:
    f = e^(4x) - sin(y^2) - sqrt(xy)

    f_(x) = 4e^(4x) - (1/2)*(1/sqrt(xy))*y

    f_(x) = 4e^(4x) - (1/2)sqrt(y/x)

    To proceed it is simpler to look at this as:
    f_(x) = 4e^(4x) - (1/2)sqrt(y)*x^(-1/2)

    So:
    f_(xx) = 16e^(4x) - (1/2)sqrt(y)*(-1/2)x^(-3/2)

    Which is the same as:
    f_(xx) = 16e^(4x) + (1/4)sqrt(y/x^3)

    The other two are done the same way, though they require a bit more work.

    -Dan
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    Quote Originally Posted by topsquark View Post
    I'll do f_(xx), you can do the rest.

    To get a partial derivative with respect to x, treat all other variables as constants. So given:
    f = e^(4x) - sin(y^2) - sqrt(xy)

    f_(x) = 4e^(4x) - (1/2)*(1/sqrt(xy))*y

    f_(x) = 4e^(4x) - (1/2)sqrt(y/x)

    To proceed it is simpler to look at this as:
    f_(x) = 4e^(4x) - (1/2)sqrt(y)*x^(-1/2)

    So:
    f_(xx) = 16e^(4x) - (1/2)sqrt(y)*(-1/2)x^(-3/2)

    Which is the same as:
    f_(xx) = 16e^(4x) + (1/4)sqrt(y/x^3)

    The other two are done the same way, though they require a bit more work.

    -Dan
    Hmm, okay, so to do f_(xy), then, we'd take f_(x), which was:

    4e^(4x) - (1/2)sqrt(y)*x^(-1/2), and then find the derivative with respect to y, so then the answer is:

    0 + {derivative of -(1/2)sqrt(y)*x^(-1/2), treating x^(-1/2) as a constant, here}

    So: -1/[4*sqrt(y)*sqrt(x)] ?

    And then, to find f_(yyx) can I just find the derivative with respect to y of the answer above and that'll be f_(yyx)?

    EDIT: hmm if you do that, you'd get 1/(8*y^(3/2)*sqrt(x)) ...
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ideasman View Post
    Hmm, okay, so to do f_(xy), then, we'd take f_(x), which was:

    4e^(4x) - (1/2)sqrt(y)*x^(-1/2), and then find the derivative with respect to y, so then the answer is:

    0 + {derivative of -(1/2)sqrt(y)*x^(-1/2), treating x^(-1/2) as a constant, here}

    So: -1/[4*sqrt(y)*sqrt(x)] ?

    And then, to find f_(yyx) can I just find the derivative with respect to y of the answer above and that'll be f_(yyx)?
    This is correct. The f_(yyx) takes a bit of simplification to get it to the form your book has.

    -Dan
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