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:
\(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds \dfrac {\begin {vmatrix} y u & u \\ v & 1 \end {vmatrix} } {\begin {vmatrix} 1 & u \\ v & 1 \end {vmatrix} }\) | \(\ds = \dfrac {u \paren {y - v} } {1 - u v}\) | |||||||||||
\(\ds \dfrac {\partial v} {\partial x}\) | \(=\) | \(\ds \dfrac {\begin {vmatrix} 1 & y u \\ v & v \end {vmatrix} } {\begin {vmatrix} 1 & u \\ v & 1 \end {vmatrix} }\) | \(\ds = \dfrac {v \paren {1 - y u} } {1 - u v}\) |
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 |
$\blacksquare$
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: $1.2$ Implicit Functions: Example $\text C$