How to deal with this implicit functions system?

Question

I have not dealt with practice of implicit functions systems yet, I just do not know how to approach it properly, so do not judge much.

The system:

$$\begin{cases} xe^{u+v} + 2uv = 1 \\ ye^{u-v} - \dfrac{u}{1+v} = 2x \end{cases}$$

defines functions $u =u(x,y)$ and $v = v(x,y)$ so $u(1,2) = 0$ and $v(1,2) = 0$

Have to find $du(1,2)$ and $dv(1,2)$

Answer 1

We need to end up with forms of equations of the following for $du$ and $dv$:

$$\frac{d}{dv}=\frac{d}{dx}\frac{\partial{x}}{\partial{v}}+\frac{d}{dy}\frac{\partial{y}}{\partial{v}}$$

$$\frac{d}{du}=\frac{d}{dx}\frac{\partial{x}}{\partial{u}}+\frac{d}{dy}\frac{\partial{y}}{\partial{u}}$$

To do this you can

1. rearrange the equations in question making $x$, and $y$ the subject.
2. Partially differentiate x and y wrt u and v, having them in terms of u and v only.
3. When $(x,y)=(1,2)$, we are given that $(u,v)=(0,0)$. Hence we can evaluate the partials in step 3 at (0,0).
4. Substitute the evaluated partials in the equations for $\frac{d}{dv}$ and $\frac{d}{du}$

I got:

$$\frac{d}{dv}=-\frac{d}{dx}$$ $$\frac{d}{du}=-\frac{d}{dx}-2\frac{d}{dy}$$

Or:

$$dv=-dx$$ $$du=-\frac{1}{\frac{d}{dx}-2\frac{d}{dy}}$$