Stuck on partial derivatives

Question

I'm stuck on a very simple task.

$$ \left\{ \begin{array}{} u = x^2-y^2\\ v=2xy \end{array} \right. $$ 1) (Solved) Calculate $\frac{\partial u}{\partial x}(1,1)$

$$\frac{\partial u}{\partial x} = 2x, \frac{\partial u}{\partial x}(1,1) = 2$$

2) (Stuck) Calculate $\frac{\partial x}{\partial u}(0,2)$

I looked at a rule: $\frac{d(u)}{d(x)} = \frac{1}{\frac{d(x)}{d(u)}}$ but that I believe that theorem was just to give the relations between the functional determinants which is not the case here.

I have really tried to check my book so what have I not understood?

By the way the correct answer is $\frac{\partial x}{\partial u}(0,2) = \frac{1}{4}$

Answer 1

The problem statement should maybe have specified that $\partial_x u$ must be evaluated at $(x,y) = (1, 1)$ in the first case, and that $\partial_u x$ must be evaluated at $(x,y) = (2, 0)$ in the second case -- or alternatively at something like $(u,v) = (4,0)$. Whatever is asked here, we have $$ u \pm \text i v = x^2 \pm 2\text i xy - y^2 = (x\pm \text i y)^2 $$ $$ \text{and}\qquad x\pm \text i y = (u \pm \text i v)^{1/2} . $$ Thus $ \frac{\partial}{\partial u} (x\pm \text i y) = \tfrac12 (u \pm \text i v)^{-1/2} $. The linear average of the $+$ and $-$ equations gives \begin{aligned} \tfrac{\partial}{\partial u} x &= \tfrac14 \big[ (u + \text i v)^{-1/2} + (u - \text i v)^{-1/2} \big] \\ &= \tfrac14 \big[ (x + \text i y)^{-1} + (x - \text i y)^{-1} \big]\\ &= \tfrac12\tfrac{x}{x^2+y^2} \, . \end{aligned} If you manage to find the missing information in the problem statement, then you can conclude.

Answer 2

It is not entirely clear, but perhaps the second question should be understood as $\frac{\partial x}{\partial u}$ at $(u,v)=(0,2)$. For this we need to treat $x$ as a function of $u$ and $v$. We get:

$$x^2=u+y^2=u+\frac{v^2}{4x^2}$$ $$x^4-ux^2-\frac{v^2}{4}=0$$ $$4x^3\frac{\partial x}{\partial u}-x^2-2ux\frac{\partial x}{\partial u}=0$$ $$\frac{\partial x}{\partial u}=\frac{x^2}{4x^3-2ux}$$

At $(u,v)=(0,2)$, we have $x^2=y^2$ and $xy=1$, so $x=\pm 1$. Therefore $$\frac{\partial x}{\partial u}(0,2)=\pm\frac14$$

which is almost what you want. Did the question say anything about whether $x$ had to be positive?