Find partial derivatives of $u=x+y+z$, $v=x^2+y^2+z^2$ and $w=x^3+y^3+z^3$

Question

I've been trying to solve this question using the Implicit functions theorem from Schaum's outline series (Theory and Problems of Differential and Integral Calculus, by Frank Ayres) with no luck:

Given $u=x+y+z$, $v=x^2+y^2+z^2$ and $w=x^3+y^3+z^3$, show that: $$\frac{\partial x}{\partial u}=\frac{yz}{(x-y)(x-z)}, \;\;\frac{\partial y}{\partial v}=\frac{y+z}{2(x-y)(y-z)}, \;\; \frac{\partial z}{\partial w}=\frac{1}{3(x-z)(y-z)}$$

The question is pretty vague, but I suppose one can assume continuity of the partial derivatives as well as differntiability whenever necessary.

Answer 1

What you need to do is, take the partial derivative of each of those equations with respect to, say, $u$. That will give you three equations, in three unknowns, ($\frac{\partial x}{\partial u}$, $\frac{\partial y}{\partial u}$, and $\frac{\partial z}{\partial u}$). Then, solve those equations for $\frac{\partial x}{\partial u}$.

Answer 2

HINT What the author means is $\frac{\partial x} {\partial u}$ holding $v$ and $w$ constant; similarly for the other two questions.

Answer 3

We better start thinking about the situation at stake here. We are given a map $${\bf f}:\quad{\bf x}\mapsto{\bf u}:={\bf f}({\bf x})\ ,$$ whereby we have written ${\bf x}:=(x,y,z)$, ${\bf u}:=(u,v,w)$. This ${\bf f}$ maps a suitably chosen region $\Omega$ in ${\bf x}$-space diffeomorphically onto some region $\Omega'$ in ${\bf u}$-space, so that there is an inverse map $${\bf g}:={\bf f}^{-1}:\quad\Omega'\to\Omega,\qquad {\bf u}\mapsto{\bf x}:={\bf g}({\bf u})\ .$$ According to a main theorem of multivariable calculus one has $$d{\bf g}({\bf u})=\left(d{\bf f}\bigl({\bf g}({\bf u})\bigr)\right)^{-1}\ .$$ The problem wants this $d{\bf g}({\bf u})$ expressed not in terms of ${\bf u}$, but in terms of the point ${\bf x}$ corresponding to ${\bf u}$. This means that we just have to compute the inverse of the Jacobian matrix $$[d{\bf f}({\bf x})]=\left[\matrix{1&1&1\cr 2x&2y&2z\cr 3x^2&3y^2&3z^2\cr}\right]\ .$$ Mathematica obtains $$[d{\bf f}({\bf x})]=\left[\matrix{{yz\over(x-y)(x-z)}&-{y+z\over 2(x-y)(x-z)}&{1\over 3(x-y)(x-z)}\cr -{xz\over(x-y)(y-z)}&{x+z\over 2(x-y)(y-z)}&-{1\over 3(x-y)(y-z)}\cr {xy\over(x-z)(y-z)}&-{x+y\over 2(x-z)(y-z)}&{1\over 3(x-z)(y-z)}\cr}\right]\ .$$ The desired partial derivatives are the three elements in the main diagonal of this matrix.