Third Partial Derivative/Examples/u = ln (x^2 + y)

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Examples of Third Partial Derivatives

Let $u = \map \ln {x^2 + y}$ be a real function of $2$ variables such that $x^2 + y \in \R_{>0}$.

Then:

$\dfrac {\partial^3 u} {\partial y^2 \partial x} = \dfrac {\partial^3 u} {\partial x \partial y^2} = \dfrac {\partial^3 u} {\partial x \partial y \partial x} = \dfrac {4 x} {\paren {x^2 + y}^3}$


Proof

\(\displaystyle \dfrac {\partial u} {\partial x}\) \(=\) \(\displaystyle \dfrac 1 {x^2 + y} \cdot 2 x\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {2 x} {x^2 + y}\)
\(\displaystyle \dfrac {\partial u} {\partial y}\) \(=\) \(\displaystyle \dfrac 1 {x^2 + y} \cdot 1\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {x^2 + y}\)
\(\displaystyle \dfrac {\partial^2 u} {\partial y \partial x}\) \(=\) \(\displaystyle \map {\dfrac \partial {\partial y} } {\dfrac {2 x} {x^2 + y} }\)
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac {2 x} {\paren {x^2 + y}^2}\)
\(\displaystyle \dfrac {\partial^2 u} {\partial x \partial y}\) \(=\) \(\displaystyle \map {\dfrac \partial {\partial x} } {\dfrac 1 {x^2 + y} }\)
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac {2 x} {\paren {x^2 + y}^2}\)
\(\displaystyle \dfrac {\partial^2 u} {\partial y^2}\) \(=\) \(\displaystyle \map {\dfrac \partial {\partial y} } {\dfrac 1 {x^2 + y} }\)
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac 1 {\paren {x^2 + y}^2}\)
\(\displaystyle \dfrac {\partial^3 u} {\partial x \partial y^2}\) \(=\) \(\displaystyle \map {\dfrac \partial {\partial x} } {-\dfrac 1 {\paren {x^2 + y}^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle -2 x \paren {-\dfrac 2 {\paren {x^2 + y}^3} }\)
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac {4 x} {\paren {x^2 + y}^3}\)
\(\displaystyle \dfrac {\partial^3 u} {\partial y \partial x \partial y}\) \(=\) \(\displaystyle \map {\dfrac \partial {\partial y} } {-\dfrac {2 x} {\paren {x^2 + y}^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle -2 x \paren {-\dfrac 2 {\paren {x^2 + y}^3} }\)
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac {4 x} {\paren {x^2 + y}^3}\)
\(\displaystyle \dfrac {\partial^3 u} {\partial y^2 \partial x}\) \(=\) \(\displaystyle \map {\dfrac \partial {\partial y} } {-\dfrac {2 x} {\paren {x^2 + y}^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle -2 x \paren {-\dfrac 2 {\paren {x^2 + y}^3} }\)
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac {4 x} {\paren {x^2 + y}^3}\)

$\blacksquare$


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