If $F(x+y+z, x^2+y^2+z^2)=0$ then find $\frac{\partial^2 z}{\partial x \partial y }$

Question

If $F(x+y+z, x^2+y^2+z^2)=0$ then find $\frac{\partial^2 z}{\partial x \partial y }$.

Attempt

I think that here we must apply the implicit differentiation Theorem, but I´dont know how I should do it, first I try use $X=x+y+z$ and $Y=x^2+y^2+z^2$ and then In case of be useful calculate $$\frac{\partial X}{\partial x}=1, \, \frac{\partial X}{\partial y}=1, \, \frac{\partial X}{\partial z}=1$$ and also $$ \frac{\partial Y}{\partial x}=2x, \, \frac{\partial Y}{\partial y}=2y, \, \frac{\partial Y}{\partial z}=2z$$ and hence my Function should looks as $$F(X,Y)=0$$ From here I Try apply the implicit function theorem which states that I should find a function $z(X)$ such that $F(X,Z(X))=0$ and that in fact $z$ is differentiable with differential equal to $$\frac{\partial z}{\partial x}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}$$ But I´m not sure about if the form of I use actually is valid, and in other case someone can clarify what is the answer(step by step (because i´m learning Analysis by myself) and more important how I should apply and understand this famous theorem.

Answer 1

Apparently, it is required to express $\dfrac{\partial^2 z}{\partial x \partial y } $ in terms of the derivatives $\dfrac{\partial F}{\partial X}$, $\dfrac{\partial F}{\partial Y}$, $\dfrac{\partial^2 F}{\partial X^2}$ etc. We will assume that all derivatives under consideration are continuous, so that the order of differentiation does not matter.

According to the implicit function theorem $$ \frac{\partial z}{\partial x}=-\left.\frac{\partial F(x+y+z, x^2+y^2+z^2)}{\partial x}\right/\frac{\partial F(x+y+z, x^2+y^2+z^2)}{\partial z}. $$ Using the chain rule, one obtains $$ \frac{\partial F(x+y+z, x^2+y^2+z^2)}{\partial x}=\frac{\partial F}{\partial X} \frac{\partial X}{\partial x}+\frac{\partial F}{\partial Y}\frac{\partial Y}{\partial x}=\frac{\partial F}{\partial X}+2x\frac{\partial F}{\partial Y} $$ $$ \frac{\partial F(x+y+z, x^2+y^2+z^2)}{\partial z}=\frac{\partial F}{\partial X} \frac{\partial X}{\partial z}+\frac{\partial F}{\partial Y}\frac{\partial Y}{\partial z}=\frac{\partial F}{\partial X}+2z\frac{\partial F}{\partial Y}, $$ (hereinafter, we mean by $\frac{\partial F}{\partial X}$ $\frac{\partial F}{\partial X}(x+y+z, x^2+y^2+z^2)$ etc.) so $$ \frac{\partial z}{\partial x}=-\frac{\dfrac{\partial F}{\partial X}+2x\dfrac{\partial F}{\partial Y}}{\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}}. $$ Let us differentiate this with respect to $y$. According to the quotient rule $$\tag{1} \frac{\partial^2 z}{\partial x\partial y} =- \frac{ \dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial X}+2x\dfrac{\partial F}{\partial Y}\right)\left(\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}\right)- \left( \dfrac{\partial F}{\partial X}+2x\dfrac{\partial F}{\partial Y}\right)\dfrac{\partial}{\partial y}\left(\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}\right) } { \left(\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}\right)^2}. $$ Let's calculate $\dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial X}\right)$ and $\dfrac{\partial}{\partial y}\left(\dfrac{\partial F}{\partial Y}\right)$ using the chain rule: $$ \dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial X}\right)= \dfrac{\partial}{\partial X}\left( \dfrac{\partial F}{\partial X}\right) \frac{\partial X}{\partial y}+ \dfrac{\partial}{\partial Y}\left( \dfrac{\partial F}{\partial X}\right) \frac{\partial Y}{\partial y}= \dfrac{\partial^2 F}{\partial X^2}+2y\dfrac{\partial^2 F}{\partial Y\partial X} $$ $$ \dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial Y}\right)= \dfrac{\partial}{\partial X}\left( \dfrac{\partial F}{\partial Y}\right) \frac{\partial X}{\partial y}+ \dfrac{\partial}{\partial Y}\left( \dfrac{\partial F}{\partial Y}\right) \frac{\partial Y}{\partial y}= \dfrac{\partial^2 F}{\partial X \partial Y}+2y\dfrac{\partial^2 F}{\partial Y^2}. $$ This implies that $$ \dfrac{\partial}{\partial y}\left( \dfrac{\partial F}{\partial X}+2x\dfrac{\partial F}{\partial Y}\right)= \dfrac{\partial^2 F}{\partial X^2}+2y\dfrac{\partial^2 F}{\partial Y\partial X}+ 2x\left( \dfrac{\partial^2 F}{\partial X \partial Y}+2y\dfrac{\partial^2 F}{\partial Y^2} \right) $$ and $$ \dfrac{\partial}{\partial y}\left(\dfrac{\partial F}{\partial X}+2z\dfrac{\partial F}{\partial Y}\right)=\dfrac{\partial^2 F}{\partial X^2}+2y\dfrac{\partial^2 F}{\partial Y\partial X}+ 2\dfrac{\partial z}{\partial y}\dfrac{\partial F}{\partial Y} + 2z\left( \dfrac{\partial^2 F}{\partial X \partial Y}+2y\dfrac{\partial^2 F}{\partial Y^2} \right). $$ It remains to calculate $\dfrac{\partial z}{\partial y}$ in the same way as before we calculated $\dfrac{\partial z}{\partial x}$, and substitute all this into (1).