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Math Help - Partial Derivative

  1. #1
    Newbie Capnrawr's Avatar
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    Partial Derivative

    Hey guys here's the problem...

    Suppose f is a differentiable function of x and y, and g(u,v) = f(e^u + sinv, e^u + cosv). Use the table of values to calculate gu(0,0) and gv(0,0).

    At (0,0): f = 3, g = 6, fx = 4, fy = 8
    At (1,2): f = 6, g = 3, fx = 2, fy = 5

    Where fx and fy and f sub x and f sub y (partial derivatives).

    My problem is, I can't figure out how to differentiate g(u,v) with respect to x or y. Any insight?
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  2. #2
    ImPerfectHacker
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    Sorry I don't know how to do this either. I guess no one knows.
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Capnrawr View Post
    Hey guys here's the problem...

    Suppose f is a differentiable function of x and y, and g(u,v) = f(e^u + sinv, e^u + cosv). Use the table of values to calculate gu(0,0) and gv(0,0).

    At (0,0): f = 3, g = 6, fx = 4, fy = 8
    At (1,2): f = 6, g = 3, fx = 2, fy = 5

    Where fx and fy and f sub x and f sub y (partial derivatives).

    My problem is, I can't figure out how to differentiate g(u,v) with respect to x or y. Any insight?
    I want to be sure I understand which variables you are refering to and "what" you are refering to with certain terms:

    I assume gu is the partial derivative of g with respect to u and gv is the same with respect to v.

    I assume that the initial values (0,0) and (1,2) refer to the initial values of u and v, such that:
    (u0,v0) = (0,0): f(u0,v0) = 3, g(u0,v0) = 6, fx(u0,v0) = 4, fy(u0,v0) = 8
    (u1,v1) = (1,2): f(u1,v1) = 6, g(u1,v1) = 3, fx(u1,v1) = 2, fy(u1,v1) = 5

    If my assumptions are correct, then I can do the problem:

    Since f(x,y) is a function of x and y,
    Where g(u,v) = f(e^u + sinv, e^u + cosv),
    It holds that
    x = e^u + sinv
    y = e^u + cosv

    Taking partial derivatives in terms of u, we get:
    Du[g(u,v)] = Du[f(e^u + sinv, e^u + cosv)] = df/dx*dx/du + df/dy*dy/du

    Notice that dx/du = e^u and dy/du = e^u, so
    gu(u,v) = e^u*fx(u,v) + e^u*fy(u,v)
    gu(0,0) = e^0*4 + e^0*8 = 4 + 8 = 12

    Taking partial derivatives in terms of v, we get:
    Dv[g(u,v)] = Dv[f(e^u + sinv, e^u + cosv)] = df/dx*dx/dv + df/dy*dy/dv

    Notice that dx/dv = cosv and dy/dv = -sinv, so
    gv(u,v) = cosv*fx(u,v) - sinv*fy(u,v)
    gv(0,0) = cos0*4 - sin0*8 = 4

    Edit: VERY BIG NOTE: the "d"s should be read as partial derivatives as all functions are in terms of more than one variable.
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  4. #4
    Newbie Capnrawr's Avatar
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    Thank you sooo much
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  5. #5
    Forum Admin topsquark's Avatar
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    I don't know what the posts were, but in the future please try to keep things professional!

    -Dan
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    I don't know what the posts were, but in the future please try to keep things professional!

    -Dan
    what are you refering to topsquark?
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  7. #7
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ImPerfectHacker View Post
    Sorry I don't know how to do this either. I guess no one knows.
    Who is the ImPerfectHacker? If this is ThePerfectHacker or CaptainBlack then I doubt there is much reason to care that this handle was used, but if not then the use of this handle is a clear violation of forum rules.

    Edit: Upon reviewing the forum rules, it seems mocking a forum moderator's handle is not a "clear violation of forum rules" but it should be.
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  8. #8
    Global Moderator

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    Quote Originally Posted by topsquark View Post
    I don't know what the posts were, but in the future please try to keep things professional!
    I think I should be more strict with the policy to post only math related stuff. No chat conversations. The Talk Spam Section is there for that.

    But this was something else, a lot of begging and another evil move.

    Who is the ImPerfectHacker? If this is ThePerfectHacker or CaptainBlack then I doubt there is much reason to care that this handle was used, but if not then the use of this handle is a clear violation of forum rules.

    Edit: Upon reviewing the forum rules, it seems mocking a forum moderator's handle is not a "clear violation of forum rules" but it should be.
    ImPerfectHacker is my good twin brother No seriously, it was a member having trying to get my attention. (Original Poster).
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by ImPerfectHacker View Post
    Sorry I don't know how to do this either. I guess no one knows.
    However annoying they may be do not create false ID's that look
    like another members to post something insulting/offensive/etc.

    Do it again and I will ban you permanently

    RonL

    (If anyone does it again I will ban them out of hand)
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