# Partial Derivative/Examples/v + ln u = x y, u + ln v = x - y

## Example of Partial Derivative

Consider the simultaneous equations:

$\begin {cases} v + \ln u = x y \\ u + \ln v = x - y \end {cases}$

Then:

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

### Second Partial Derivative

$\dfrac {\partial^2 u} {\partial x^2} = \dfrac {u \paren {\paren {y - v}^2 - v \paren {\paren {1 - y u} + u \dfrac {y - v} {1 - u v} \paren {1 + y - v - y u} } } } {\paren {1 - u v}^2}$

## Proof

 $\ds v + \ln u$ $=$ $\ds x y$ $\ds \leadsto \ \$ $\ds \map {\dfrac \partial {\partial x} } {v + \ln u}$ $=$ $\ds \map {\dfrac \partial {\partial x} } {x y}$ $\ds \leadsto \ \$ $\ds \dfrac {\partial v} {\partial x} + \dfrac 1 u \dfrac {\partial u} {\partial x}$ $=$ $\ds y$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds \dfrac {\partial u} {\partial x} + u \dfrac {\partial v} {\partial x}$ $=$ $\ds y u$

 $\ds u + \ln v$ $=$ $\ds x - y$ $\ds \leadsto \ \$ $\ds \map {\dfrac \partial {\partial x} } {u + \ln v}$ $=$ $\ds \map {\dfrac \partial {\partial x} } {x - y}$ $\ds \leadsto \ \$ $\ds \dfrac {\partial u} {\partial x} + \dfrac 1 v \dfrac {\partial v} {\partial x}$ $=$ $\ds 1$ $\text {(2)}: \quad$ $\ds \leadsto \ \$ $\ds v \dfrac {\partial u} {\partial x} + \dfrac {\partial v} {\partial x}$ $=$ $\ds v$

and so combining $(1)$ and $(2)$ into matrix form:

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

Hence by Cramer's Rule:

 $\ds \dfrac {\partial u} {\partial x}$ $=$ $\ds \dfrac {\begin {vmatrix} y u & u \\ v & 1 \end {vmatrix} } {\begin {vmatrix} 1 & u \\ v & 1 \end {vmatrix} }$ $\ds$ $=$ $\ds \dfrac {u \paren {y - v} } {1 - u v}$ Definition of Determinant

and:

 $\ds \dfrac {\partial v} {\partial x}$ $=$ $\ds \dfrac {\begin {vmatrix} 1 & y u \\ v & v \end {vmatrix} } {\begin {vmatrix} 1 & u \\ v & 1 \end {vmatrix} }$ $\ds$ $=$ $\ds \dfrac {v \paren {1 - y u} } {1 - u v}$ Definition of Determinant

$\blacksquare$