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Math Help - Partial Derivatives

  1. #1
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    Partial Derivatives

    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 fx=8

    I have no clue how to do this someone please help
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  2. #2
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    Quote Originally Posted by multivariablecalc View Post
    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 fx=8

    I have no clue how to do this someone please help
    Let x = e^u+\sin v and y = e^u+\cos v so

     <br />
g_u = f_x x_u + f_y y_u = e^u f_x + e^u f_y<br />
     <br />
g_v = f_x x_v + f_y y_v = \cos v f_x - \sin v f_y<br />

    Do you have a table of values for f_x(1,2) and f_y(1,2)?
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  3. #3
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    If the problem say "use the table of values" you had better have a table of values! Please post the entire problem, including the table of values.
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  4. #4
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    fx(1,2)=2, fy(1,2)=5, f(0,0)=3, f(1,2)=6, g(0,0)=6, g(1,2)=3, fx(0,0)=4,fy(0,0)=8
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  5. #5
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    Welcome back! Do you see why you needed to know fx(1,2) and fy(1,2)?
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  6. #6
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    Well I am not the same person who posted this thread. However, assuming this person was referring to problem #11 of section 11.5 of Essential Calculus-Early Transcendentals (James Stewart), then these are the correct values. If I am not mistaken, gu(0,0)=fx(1,2)e^u +fy(1,2)e^u =2e^0 +5e^0=7. gv(0,0) is not much different.
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  7. #7
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    Oh, I see! (Although if you had gone on a 4-month quest for the table of values perhaps you would have have returned a different person!)

    I don't think you're mistaken. Hopefully you can also see how Danny was holding one variable constant while differentiating with respect to the other.
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    Re: Partial Derivatives

    Hopefully bringing this thread back to life.

    I'm working on the exact same problem, and I have a handful of questions.

    First of all, What exactly is G(v) and G(u)? I'm trying to follow along on this, and I'm understanding most other questions around it, but I'm at a loss for this one.

    I understand transfering the initial equation down to x=e^u +sin(v) and y=e^u +cos(v),

    but after that, I don't quite understand how to get to G(u)=Fx(Xu)+Fy(Yu)=e^(u)Fx + e^(u)Fy, or the same for G(v)

    Any help is appreciated. Apologies for not fantastic formatting, just started up.
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  9. #9
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    Re: Partial Derivatives

    Quote Originally Posted by Sabertooth View Post
    First of all, What exactly is G(v) ...?
    No one put the v in brackets - now it reads "function G of v" but you want to be saying "derivative with respect to v of g (which is a function of u and v)." So you say gv, if you can format it so. (Maybe why you resorted to upper case g.) Or gv(u,v), or for particular values 0 and 0 for u and v, gv(0,0). In other words, dg/dv.

    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    is the chain rule for two inner functions, i.e...

    \frac{d}{du}\ f(x(u), y(u)) = \frac{\partial f}{\partial x} \frac{dx}{du} + \frac{\partial f}{\partial y} \frac{dy}{du}

    As with...



    ... the ordinary chain rule, straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the (corresponding) dashed balloon expression which is (one of) the inner function(s) of the composite expression.



    Sorry I cluttered it a bit, because of the notation issue.


    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods
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