# problem about implicit function theorem and implicit partial derivatives

Let $$x,y,u,v$$ be variables that satisfy these two equations: $$F: x+y - u-v =0$$ $$G: x+u - e^{y+v} =0$$

Show that at neighbourhood of $$(x,y,u,v) = (-\frac{1}{2} , 1,\frac{3}{2} , -1)$$ we can express $$u$$ and $$v$$ as functions of $$x$$ and $$y$$ and then find partial derivatives of $$u$$ and $$v$$ with respect to $$x$$ and $$y$$ at point $$(x,y) = (-\frac{1}{2} , 1)$$

For the first part I simply used implicit function theorem.

$$\frac{\partial(F,G)}{\partial(u,v)}$$ at given point is $$\begin{vmatrix} -1 & -1 \\ 1 & -1 \\ \end{vmatrix} = 2 \neq 0$$ So we can express $$u$$ and $$v$$ as function of $$x$$ , $$y$$

For the second part I simply assumed that $$x$$ and $$y$$ are independent and $$u$$ and $$v$$ are $$u(x,y)$$ and $$v(x,y)$$ and by calculating partial derivative from both equations I get:

$$1 - \frac{\partial u}{\partial x} - \frac{\partial v}{\partial x} = 0$$ $$1 + \frac{\partial u}{\partial x} - \frac{\partial v}{\partial x} e^{y+v} =0$$

and by putting the point $$(x,y,u,v) = (-\frac{1}{2} , 1,\frac{3}{2} , -1)$$ and solving the easy system, we get $$\frac{\partial u}{\partial x} = 0$$ and $$\frac{\partial v}{\partial x} = 1$$

Similarly by calculating this for y, we get $$\frac{\partial u}{\partial y} = 1$$ and $$\frac{\partial v}{\partial y} = 0$$

Someone told me the solution for second part is wrong and I can't solve this question like this. He said that you must reason it very well if you want to solve this way and show that $$u$$ and $$v$$ are really functions of $$x$$ and $$y$$! But I think I showed this in the first part! Also I solved this using usual method that involves matrices and their inverses and I got the same answers.

Now, I want to make sure whether my method is correct or not? And if it is correct, how can I reason that my way is correct. Note that I encountered this question in an undergraduate Calculus-2 exam.

For convenience of notation, denote $$\alpha := \left( -\frac{1}{2}, 1 \right)$$. By the Implicit Function Theorem, there exists an open neighbourhood $$A \subset \Bbb{R}^2$$ of the point $$\alpha$$, and unique ($$C^{\infty}$$) functions $$u:A \to \Bbb{R}$$, and $$v:A \to \Bbb{R}$$ such that \begin{align} u(\alpha) = \frac{3}{2} \quad \text{and} \quad v(\alpha) = 1, \end{align} and for all $$(x,y) \in A$$, we have \begin{align} \begin{cases} x + y - u(x,y) - v(x,y) &= 0 \\\\ x + u(x,y) - e^{y + v(x,y)} &= 0. \end{cases} \end{align}
Differentiating this equation and computing the derivative at $$\alpha$$ implies that \begin{align} \begin{cases} 1 - \dfrac{\partial u}{\partial x}(\alpha) - \dfrac{\partial v}{\partial x}(\alpha) &= 0 \\\\ 1 + \dfrac{\partial u}{\partial x}(\alpha) - \dfrac{\partial v}{\partial x}(\alpha) &= 0 \end{cases} \end{align} Hence, \begin{align} \dfrac{\partial u}{\partial x}(\alpha) = 0 \quad \text{and} \quad \dfrac{\partial v}{\partial x} (\alpha) = 1. \end{align}
The only difference between what I wrote and what you wrote is that I was very explicit in mentioning the domain and target space of the functions $$u,v$$, I was careful with my choice of words regarding logical quantifiers (i.e "for all", and "there exists") and I explicitly wrote down where all the derivatives are being computed.