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Math Help - derivatives

  1. #1
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    Dec 2011
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    derivatives

    Hello,

    I got the following problems and I am not sure how to do them:

    fx(x,y) if f(x.y) = 4x-3y+6

    fy(x,y) if f(x.y) = 3x^2+2xy-7y^2

    and find fx(x.y) and fy(x.y) f(x.y) = square root(2x-y^2)

    fxx(x.y) if f(x.y) = 6x-5y+3

    and

    dz/dx if z = x^3+4x^2y+2y3^3

    Please help
    Last edited by silver163; April 19th 2012 at 07:30 AM.
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  2. #2
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    Re: derivatives

    fsubx is the partial derivative of f with respect to x. So take y as a constant in the f(x,y) expression before differentiating.

    Example:
    f(x,y)=4x-3y+6
    fx(x,y)=4

    Similarly, fsuby is the partial derivative with respect to y. Now you will treat x as a constant.
    Example:
    f(x,y)=3x^2+2xy-7y^2
    fy(x,y)=2x-14y

    Finally, fxx is the second partial derivative. Just follow the procedure above a second time, with y held constant again.

    Hope that helps,
    Cheerios
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  3. #3
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    Re: derivatives

    but if y is a constant it would go to 0 when derived no? same for x but you still have x in the final answer.
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  4. #4
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    Re: derivatives

    You must use the product rule. From single-variable calculus:
    Given y(x)=cx
    y'(x)=c[x(d/dx)]+x[c(d/dx)]=c*1+x*0=c

    Look at the term 2xy in the second example function above. The partial derivative with respect to y looks like:
    y[2x(d/dy)]+2x[y(d/dy)]=y*0+2x*1=2x
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