Second Partial Problem

[Trenters4325]

Let g be a function that maps vectors in
R^3
to vectors in
R^1
and \(\displaystyle $$
{{\partial g} \over {\partial x}}is not equal to
0
$$\). By the IF T, g=0 defines z implicitly as a function \(\displaystyle $$
\phi
$$\) of x and y. Assume g is second-order continuously differentiable. Show that \(\displaystyle $$
{{\partial ^2 z} \over {\partial x^2 }} = {{{{\partial ^2 f} \over {\partial z^2 }}\left( {{{\partial f} \over {\partial x}}} \right)^2 - 2{{\partial ^2 f} \over {\partial x\partial z}}{{\partial f} \over {\partial x}}{{\partial f} \over {\partial z}} + {{\partial ^2 f} \over {\partial x^2 }}\left( {{{\partial f^{} } \over {\partial z}}} \right)^2 } \over {\left( {{{\partial f} \over {\partial z}}} \right)^3 }}
$$\).

I tried partially differentiating \(\displaystyle $$
{{\partial z} \over {\partial x}} = {{ - {{\partial f} \over {\partial x}}} \over {{{\partial f} \over {\partial z}}}}
$$\) with respect to x using the quotient rule but the result looks drastically different than above. Obviously, the denominator is only squared with the quotient rule.

Sorry the TeX got messed up. I still think its understandable though.

 

[Trenters4325]

Let me know if its not understandable.