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Thread: f"(x)

  1. #1
    Jul 2007


    Letís F be a function of two variables (x,y) which is continuous in a neighborhood of (x0,y0) while F(x0,y0) = 0 and suppose it would have continuous partial derivatives Fx ,Fy ,Fxx ,Fxy ,Fyy,Fyx in this neighborhood and Fy(x0,y0) # 0 .
    Prove that F(x,y) has a unique answer as function f in a neighborhood of x0 such that F(x,f(x)) = 0 and f is continuously two times differentiable by the following formula :

    fĒ(x) = - (Fxx(x,y)( Fy(x,y))^2 - 2 Fx(x,y) Fxy(x,y)+ Fyy(x,y)( Fx(x,y))^2 )/ (Fy(x,y))^3
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    The existence and differentiability of f come from the implicit function theorem.

    To obtain the formula for f'', you need to use the chain rule in a crafty way. Write $\displaystyle y=f(x)$. Then y satisfies the equation $\displaystyle F(x,y)=0$. Now regard F as a function of the two variables x and y, each of which is in turn a function of x (via the equations x=x and y=f(x)). Differentiate the equation $\displaystyle F(x,y)=0$ with respect to x, using the chain rule, and you get $\displaystyle F_x(x,y) + f'(x)F_y(x,y) = 0$. Now repeat the process, differentiating that equation with respect to x. This gives $\displaystyle F_{xx}(x,y) + f'(x)F_{xy}(x,y) + f''(x)\bigl(F_{yx}(x,y) + f'(x)F_{yy}(x,y)\bigr) = 0$.

    Solve the first of those two equations to get an expression for f'(x). Substitute this into the second equation and you'll get the given expression for f''(x). (You'll need to quote the theorem which says that $\displaystyle F_{yx} = F_{xy}$ provided that these both exist and are continuous.)
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