Implicit function problem

Question

maybe I'm just blind but I cannot see what to do here: $$F(x_1+x_2+x_3,x_1^2+x_2^2+x_3^2)=0$$ $$x_3=f(x_1,x_2)$$ I would like to obtain the differential of $f$ in terms of the partial derivatives of $F$. This screams implicit function theorem, but, well... As it does not say $F(x_1,x_2,x_3)$ I have no idea how to continue... Thank you for your hints!

Answer 1

If you are asking for partial derivatives of $\ f$, then I think the following might help.

First some convention: Let $ { g }_{ i } $ denote the partial derivative with respect to the i-th entry of the function $g$. Then, by chain rule,

$\frac { \partial F }{ \partial { x }_{ 1 } } = {F}_{1}(1+{f}_{1}) + {F}_{2}(2{x}_{1}+2 f{f}_{1}) = 0 $. We can obtain this by substituting ${x}_{3}$ by $\ f$.

Another substitution and

$\frac { \partial F }{ \partial { x }_{ 1 } } = {F}_{1}(1+{f}_{1}) + {F}_{2}(2{x}_{1}+2 {x}_{3} {f}_{1}) = 0 $

From above we can get ${f}_{1}$. Similarly, we can also get ${f}_{2}$.

Answer 2

Some thoughts about your question.

Well, you see, your F is composition of functions, and you should differentiate it as a composition:

$F(x_1 + x_2 + x_3, x_1^2 + x_2^2 + x_3^2) = F(g(x_1, x_2, x_3),h(x_1, x_2, x_3))$ where $g(x_1, x_2, x_3) = x_1 + x_2 + x_3$, and $h(x_1, x_2, x_3) = x_1^2 + x_2^2 + x_3^2$. So $dF = F'_g dg + F'_h dh$.