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Math Help - Implicit differentiation

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    Implicit differentiation

    Let F: R^2 \rightarrow R be a function C^2 for which F(x+y,y+z)=0 defines implicitly z=f(x,y) around a point (a,b,c). Verify that in (a,b)
    dz/dx-dz/dy=1

    I'm having difficulty with getting rid of some derivatives, they should cancel out I guess.

    (It's a bit heavy to read but please take a look at my work)

    First, I apply the chain rule with u=g(x,y)=x+y, v=h(x,y)

    and get the expressions for dz/dx and dz/dy

    dF/dx=dF/du dg/dx+dF/dv dh/dx=dF/du+dF/dv dz/dx

    solve for dz/dx,

    dz/dx=(dF/dx-dF/du)/(dF/dv)

    Then,

    dF/dy=dF/du+dF/du dz/dy

    dz/dy={dF/dy-dF/du}/{dF/dv}

    So dz/dx-dz/dy={dF/dx-dF/dy-2dF/du}/{dF/dv}=???

    From dF/dz=dF/dv but dF/dz is no good.

    How can I get rid of those derivatives?
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  2. #2
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    Quote Originally Posted by kodos View Post
    Let F: R^2 \rightarrow R be a function C^2 for which F(x+y,y+z)=0 defines implicitly z=f(x,y) around a point (a,b,c). Verify that in (a,b)
    dz/dx-dz/dy=1

    I'm having difficulty with getting rid of some derivatives, they should cancel out I guess.

    (It's a bit heavy to read but please take a look at my work)

    First, I apply the chain rule with u=g(x,y)=x+y, v=h(x,y)

    and get the expressions for dz/dx and dz/dy

    dF/dx=dF/du dg/dx+dF/dv dh/dx=dF/du+dF/dv dz/dx

    solve for dz/dx,

    dz/dx=(dF/dx-dF/du)/(dF/dv)

    Then,

    dF/dy=dF/du+dF/du dz/dy

    dz/dy={dF/dy-dF/du}/{dF/dv}

    So dz/dx-dz/dy={dF/dx-dF/dy-2dF/du}/{dF/dv}=???

    From dF/dz=dF/dv but dF/dz is no good.

    How can I get rid of those derivatives?

    You have an error in your \frac{dF}{dy}


    You should get

    \frac{dF}{dy}=\frac{dF}{du}\frac{du}{dy}+\frac{dF}  {dv}\frac{dv}{dy}+\frac{dF}{dv}\frac{dv}{dz}\frac{  dz}{dy}=0
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