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:

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

Proof

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

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

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:

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

and:

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

$\blacksquare$