Partial Derivative/Examples/v + ln u = x y, u + ln v = x - y/Second Partial Derivative
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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:
- $\dfrac {\partial^2 u} {\partial x^2} = \dfrac {u \paren {\paren {y - v}^2 - v \paren {\paren {1 - y u} + u \dfrac {y - v} {1 - u v} \paren {1 + y - v - y u} } } } {\paren {1 - u v}^2}$
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Proof
We have from Partial Derivatives of $v + \ln u = x y$, $u + \ln v = x - y$ that:
\(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds \dfrac {u \paren {y - v} } {1 - u v}\) | ||||||||||||
\(\ds \dfrac {\partial v} {\partial x}\) | \(=\) | \(\ds \dfrac {v \paren {1 - y u} } {1 - u v}\) |
Hence:
\(\ds \dfrac {\partial^2 u} {\partial x^2}\) | \(=\) | \(\ds \map {\dfrac \partial {\partial x} } {\dfrac {u \paren {y - v} } {1 - u v} }\) | Definition of Second Partial Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {1 - u v} \map {\dfrac \partial {\partial x} } {u \paren {y - v} } - \paren {u \paren {y - v} } \map {\dfrac \partial {\partial x} } {1 - u v} } {\paren {1 - u v}^2}\) | Quotient Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {1 - u v} \paren {\dfrac {\partial u} {\partial x} \paren {y - v} + u \map {\dfrac \partial {\partial x} } {y - v} } - \paren {u \paren {y - v} } \paren {-\paren {\dfrac {\partial u} {\partial x} v + u \dfrac {\partial v} {\partial x} } } } {\paren {1 - u v}^2}\) | Product Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {1 - u v} \paren {\dfrac {\partial u} {\partial x} \paren {y - v} - u \dfrac {\partial v} {\partial x} } + \paren {u \paren {y - v} } \paren {\dfrac {\partial u} {\partial x} v + u \dfrac {\partial v} {\partial x} } } {\paren {1 - u v}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {1 - u v} \paren {\paren {\dfrac {u \paren {y - v} } {1 - u v} } \paren {y - v} - u \paren {\dfrac {v \paren {1 - y u} } {1 - u v} } } + \paren {u \paren {y - v} } \paren {\paren {\dfrac {u \paren {y - v} } {1 - u v} } v + u \paren {\dfrac {v \paren {1 - y u} } {1 - u v} } } } {\paren {1 - u v}^2}\) | substituting for $\dfrac {\partial u} {\partial x}$ and $\dfrac {\partial v} {\partial x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {u \paren {y - v}^2 - u v \paren {1 - y u} + u \paren {y - v} \paren {\paren {\dfrac {u v \paren {y - v} } {1 - u v} } + \paren {\dfrac {u v \paren {1 - y u} } {1 - u v} } } } {\paren {1 - u v}^2}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {u \paren {\paren {y - v}^2 - v \paren {\paren {1 - y u} + u \dfrac {y - v} {1 - u v} \paren {1 + y - v - y u} } } } {\paren {1 - u v}^2}\) | simplifying |
$\blacksquare$
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $13$