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

From ProofWiki
Jump to navigation Jump to search

Example of Partial Derivative

Consider the simultaneous equations:

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


Then:

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




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