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

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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

$\blacksquare$


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