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calculus question

  • 09-03-2011 10:58pm
    #1
    Registered Users, Registered Users 2 Posts: 603 ✭✭✭


    f(x; y) = (x - 1)(y -2)(x + y)
    Calculate fx; fy; fxx; fyy; fxy

    anyone know how to do this?
    Im guessing the first one is just get x on its own,then get y on its own for the second one.how bout the next three?just the method now the actual answer please :) thanks!


Comments

  • Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    This involves the differentiation of functions with more than one variable, two in this case. The method is called partial differentiation. You can see a bit about that here.

    The first thing you need to do is calculate the first-order derivatives, those are [latex]f_{x}[/latex] and [latex]f_y[/latex]. To calculate [latex]f_{x}[/latex] you differentiate f with respect to x and hold y constant. i.e. you just differentiate x as you normally would and treat y as if it were a constant. Then for [latex]f_y[/latex] you do the opposite, differentiate w.r.t y and hold x constant.

    An example would be if f(x,y)=2xy then [latex]f_x[/latex] would be 2y and [latex]f_y[/latex] would be 2x. Try the first order ones first and worry about the others after.


  • Registered Users, Registered Users 2 Posts: 3,038 ✭✭✭sponsoredwalk


    I'll assume that fx is notation for the partial derivative with respect to x.
    Well in f(x,y) what you could do is multiply the whole thing out & then
    take the derivative w.r.t. x holding y constant etc...

    The notation fyy would be better written as (fx)y which is just saying that
    you should take the partial derivative of (fx) w.r.t. y. If (fx) = Q then we
    can write (fx)y = Qy. You get what I'm saying hopefully :) Basically just
    take the partial derivative with respect to y of the partial derivative with
    respect to x of f(x,y). Another way to say this is that you just take the
    partial derivative with respect to y of the partial derivative of f(x,y) with
    respect to x. You could also use thee product rule on this if you wanted,
    instead of multiplying it all out.

    edit: Didn't see Red respond, I was 2 minutes too slow! :D


  • Registered Users, Registered Users 2 Posts: 13,077 ✭✭✭✭bnt


    I see they've asked you to calcuate [latex]f_{xy}[/latex] but not [latex]f_{yx}[/latex], which is probably because they should both give the same result. If you're new to this I suggest you do both anyway, and if they're different, something's gone Wrong somewhere. :p

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