# Expressing variables in terms of eachother using partial derivatives/Jacobian.

I am having some trouble with a question recently regarding partial derivatives/multivariable calculus. Here is the question:

Suppose that $$(x, y, u, v)$$ $$∈$$ $$R^4, v \ne$$ −1, satisfy equations

$$e^x + e^y + cos u + sin(2v) = 3$$

$$x + (e^2)^y + u^2 + log((v + 1)^2) = 1$$

I have to find which two variables out of $$x,y,u,v$$ can be expressed uniquely in terms of the other two near the origin such that $$(x,y,u,v) = (0,0,0,0).$$

I have taken$$z^T = (x,y,u,v)$$ and $$f(z) = \begin{pmatrix}e^x + e^y + cos(u) + sin(2v)-3\\x + (e^2)^y + u^2 + log((v+1)^2)-1\end{pmatrix} , f(z)=0$$

I then took the partial derivatives of each element, giving the Jacobian matrix to be:

$$Jf(x) = \begin{pmatrix}e^x &e^y &sin(u)&cos(2v)\\1&2e^{2y} & 2u & {2\over v+1}\end{pmatrix}$$

Giving $$Jf(0,0,0,0) = \begin{pmatrix}1&1&0&2\\1&2&0&2\end{pmatrix}$$

However I am unsure on how to proceed from here in finding which two variables can be expressed in terms of the other two. Any help would be greatly appreciated.

• Something is wrong with your Jacobian matrix. Jacobian matrix of $f \colon \mathbb{R}^d \to \mathbb{R}^k$ is $\Big( \frac{\partial f_i}{\partial x^j} \Big)_{ij}$. Dec 2, 2017 at 14:28
• Can you clarify what's wrong with it? Dec 2, 2017 at 14:30
• Well, it should be like $$\begin{pmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} & \frac{\partial f_1}{\partial u} & \frac{\partial f_1}{\partial v} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} & \frac{\partial f_2}{\partial u} & \frac{\partial f_2}{\partial v} \end{pmatrix}$$ Dec 2, 2017 at 14:39
• Oh of course, hang on it let me change it. Dec 2, 2017 at 14:40

## 1 Answer

Implicit function theorem states that if the submatrix $$\begin{pmatrix} \frac{\partial f_1}{\partial \zeta} & \frac{\partial f_1}{\partial \xi} \\ \\ \frac{\partial f_{2}}{\partial \zeta} & \frac{\partial f_2}{\partial \xi} \end{pmatrix}$$ of $Jf$ is invertible, than you can solve the equation for $\xi$ and $\zeta$. In your case the submatrix $$\begin{pmatrix} 1 & 1 \\ 1 & 2\end{pmatrix}$$ is invertible, so, the first two variables could be expressed in terms of second two.

• Thanks so much! Dec 2, 2017 at 14:58