0
$\begingroup$

Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ be defined as $$ (u, v, w)=f(x,y,z) = (e^{2y} + e^{2z}, e^{2x} - e^{2z}, x - y) $$ I have shown that it has differentiable inverse in a neighborhood of every point in $\mathbb{R}^3$. I want to compute $$ \frac{\partial x}{\partial u}, \frac{\partial x}{\partial v}, \frac{\partial x}{\partial w} $$ in terms of $x,y,z$. Based on the first part, I applied the Inverse Function Theorem and computed the inverse Jacobian $$ \partial f^{-1}(u, v, w) = \frac{1}{2(e^{2x} + e^{2y})}\begin{pmatrix} 1 & 1 & 2e^{2y} \\ 1 & 1 & -2e^{2x} \\ e^{2x - 2y} & -e^{2y - 2z} & 2e^{2x+ 2y - 2z} \end{pmatrix} $$ However, I don't know how to proceed. What is $\frac{\partial x}{\partial u}$ anyway? Is it going to be number or a vector? All I can compute is $\frac{\partial f}{\partial x}$ and $\frac{\partial f^{-1}}{\partial u}$. Also the inverse Jacobian isn't even in terms of $u, v, w$. As you can tell I am really struggling to understand multivariable differentiation so any help would be much appreciated.

$\endgroup$

1 Answer 1

0
$\begingroup$

Hint: Use $$u=e^{2y}+e^{2z}$$ $$v=e^{2x}-e^{2z}$$ $$w=x-y$$ throughout, in this manner, for example $u+v=e^{2x}+e^{2y}$, etcetera.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .