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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.

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

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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.

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  • $\begingroup$ Thanks so much! $\endgroup$ Dec 2, 2017 at 14:58

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