Construction of derivative as ratio of determinants related to simultaneous system of implicit functions

Solution 1:

Use implicit differentiation of $v + \ln u = x y$ and $u + \ln v = x - y$ to get expressions for $\dfrac {\partial u} {\partial x}$ and $\dfrac {\partial v} {\partial x}$.

You should get: $$\dfrac {\partial u} {\partial x} + u \dfrac {\partial v} {\partial x} = y u$$

and: $$v \dfrac {\partial u} {\partial x} + \dfrac {\partial v} {\partial x} = v$$

Then you can combine the above into simultaneous linear equations expressed in matrix form as:

$$\begin {pmatrix} 1 & u \\ v & 1 \end {pmatrix} \begin {pmatrix} \dfrac {\partial u} {\partial x} \\ \dfrac {\partial v} {\partial x} \end {pmatrix} = \begin {pmatrix} y u \\ v \end {pmatrix}$$

and then it's a matter of using Cramer's Rule to evaluate:

$$\dfrac {\partial u} {\partial x} = \dfrac {\begin {vmatrix} y u & u \\ v & 1 \end {vmatrix} } {\begin {vmatrix} 1 & u \\ v & 1 \end {vmatrix} } = \dfrac {u (y - v) } {1 - u v}$$

and:

$$\dfrac {\partial v} {\partial x} = \dfrac {\begin {vmatrix} 1 & y u \\ v & v \end {vmatrix} } {\begin {vmatrix} 1 & u \\ v & 1 \end {vmatrix} } = \dfrac {v (1 - y u) } {1 - u v}$$

exactly as though this were a pair of simultaneous equations.

This of course is done (either explicitly or implicitly) by taking the inverse of the matrices $\begin {pmatrix} 1 & u \\ v & 1 \end {pmatrix}$ and $\begin {pmatrix} y u & u \\ v & 1 \end {pmatrix}$ and $\begin {pmatrix} 1 & y u \\ v & v \end {pmatrix}$.

This involves calculating their determinants.