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Thread: Partial Derivatives

  1. #1
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    Partial Derivatives

    Let u = x + y and v = x -y so that x = (u + v)/2 and y = (u-v)/2
    Given a twice continuously differentiable functions f(x,y) set
    g(u,v) = f( (u+v)/2, (u-v)/2) so that f(x,y) = g( x+y, x-y)
    1) Relate ∂f/∂x, ∂f/∂y to ∂g/∂u, ∂g/∂v
    2) Relate ∂^2f/∂x^2, ∂^2f/∂y^2 to ∂^2g/∂u^2 , ∂^2g/∂v^2 and ∂^2g/∂u∂v
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  2. #2
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    Try taking the partial derivatives and see if you notice anything about them.
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  3. #3
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    how do I take those partials if they aren't defined explicitly?
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  4. #4
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    You can apply the limit definition of partial derivatives.
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    \displaystyle \frac{\partial f}{\partial x}(x,y)=\lim_{h\to 0}\frac{f((x+h),y)-f(x,y)}{h}
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  6. #6
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    well the first part I said, using the chain rule
    ∂g/∂u = ∂g/∂x ∂x/∂u + ∂g/∂y ∂y/∂u = (1/2) ∂f/∂x + (1/2) ∂f/∂y
    ∂g/∂v = ∂g/∂x ∂x/∂v + ∂g/∂y ∂y/∂∂v = (1/2) ∂f/∂x - (1/2) ∂f/∂y because f(x,y) = g(u,v)

    for the second part
    ∂^2/∂x^2 = ∂/∂x (∂g/∂u ∂u/∂x + ∂g/∂v ∂v/∂x)
    where should I go from here?
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  7. #7
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    Quote Originally Posted by Playthious View Post
    Let u = x + y and v = x -y so that x = (u + v)/2 and y = (u-v)/2
    Given a twice continuously differentiable functions f(x,y) set
    g(u,v) = f( (u+v)/2, (u-v)/2) so that f(x,y) = g( x+y, x-y)
    1) Relate ∂f/∂x, ∂f/∂y to ∂g/∂u, ∂g/∂v
    2) Relate ∂^2f/∂x^2, ∂^2f/∂y^2 to ∂^2g/∂u^2 , ∂^2g/∂v^2 and ∂^2g/∂u∂v
    \displaystyle f(x,y)=g(x+y,x-y)=g(u,v)=f\left(\frac{u+v}{2},\frac{u-v}{2}\right)

    \displaystyle \frac{\partial f}{\partial x}=\frac{\partial g}{\partial u}

    \displaystyle \frac{\partial f}{\partial y}=\frac{\partial g}{\partial v}
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