Second Partial Derivative/Examples/u + ln u = x y

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Examples of Second Partial Derivative

Let $u + \ln u = x y$ be an implicit function.

Then:

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


Proof

From Partial Derivative Examples: u + ln u = x y:

\(\displaystyle \dfrac {\partial u} {\partial x}\) \(=\) \(\displaystyle \dfrac {u y} {u + 1}\)
\(\displaystyle \dfrac {\partial u} {\partial y}\) \(=\) \(\displaystyle \dfrac {u x} {u + 1}\)


Then:

\(\displaystyle \dfrac {\partial^2 u} {\partial y \partial x}\) \(=\) \(\displaystyle \map {\dfrac \partial {\partial y} } {\dfrac {u y} {u + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\paren {u + 1} \dfrac \partial {\partial y} u y - u y \dfrac \partial {\partial y} \paren {u + 1} } {\paren {u + 1}^2}\) Quotient Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\paren {u + 1} \paren {y \dfrac {\partial u} {\partial y} + u \dfrac {\partial y} {\partial y} } - u y \dfrac \partial {\partial y} \paren {u + 1} } {\paren {u + 1}^2}\) Product Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\paren {u + 1} \paren {y \dfrac {\partial u} {\partial y} + u} } {\paren {u + 1}^2} - \dfrac {u y \dfrac {\partial u} {\partial y} } {\paren {u + 1}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\paren {u + 1} u} {\paren {u + 1}^2} + \dfrac {\paren {u + 1} y - u y } {\paren {u + 1}^2} \dfrac {\partial u} {\partial y}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac u {u + 1} + \dfrac y {\paren {u + 1}^2} \dfrac {\partial u} {\partial y}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac u {u + 1} + \dfrac y {\paren {u + 1}^2} \dfrac {u x} {u + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac u {u + 1} + \dfrac {x y u} {\paren {u + 1}^3}\)


Then we have that:

$\dfrac {\partial^2 u} {\partial x \partial y} = \map {\dfrac \partial {\partial x} } {\dfrac {u x} {u + 1} }$

which is exactly the same as:

$\dfrac {\partial^2 u} {\partial y \partial x} = \map {\dfrac \partial {\partial y} } {\dfrac {u y} {u + 1} }$

but with $y$ and $x$ exchanged.


Hence:

\(\displaystyle \dfrac {\partial^2 u} {\partial x \partial y}\) \(=\) \(\displaystyle \map {\dfrac \partial {\partial x} } {\dfrac {u x} {u + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac u {u + 1} + \dfrac {y x u} {\paren {u + 1}^3}\)

and it is seen that the second partial derivatives are indeed the same.

$\blacksquare$


Sources