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.


1 Answer 1


I think your solution is fine, but of course it all depends on how picky your marker is. It is very common in differential calculus, and especially in the context of partial derivatives and implicit function theorem to abuse notation by suppressing where the derivatives are being evaluated, and to confuse a function for its values etc. One good thing you did is to (sometimes) say in words that you are evaluating the derivatives at a given point.

To be more precise, I would phrase the argument as follows (after showing the determinant is non-zero, so that we can apply the IFT):

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.

I think it would be good to practice writing "mathematically precise" statements ... but from a conceptual standpoint, I don't see anything (deeply/fundamentally) wrong with what you have done (only minor nitpicks).


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