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

## 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

 $\ds \dfrac {\partial u} {\partial x}$ $=$ $\ds \dfrac 1 {x^2 + y} \cdot 2 x$ $\ds$ $=$ $\ds \dfrac {2 x} {x^2 + y}$ $\ds \dfrac {\partial u} {\partial y}$ $=$ $\ds \dfrac 1 {x^2 + y} \cdot 1$ $\ds$ $=$ $\ds \dfrac 1 {x^2 + y}$ $\ds \dfrac {\partial^2 u} {\partial y \partial x}$ $=$ $\ds \map {\dfrac \partial {\partial y} } {\dfrac {2 x} {x^2 + y} }$ $\ds$ $=$ $\ds -\dfrac {2 x} {\paren {x^2 + y}^2}$ $\ds \dfrac {\partial^2 u} {\partial x \partial y}$ $=$ $\ds \map {\dfrac \partial {\partial x} } {\dfrac 1 {x^2 + y} }$ $\ds$ $=$ $\ds -\dfrac {2 x} {\paren {x^2 + y}^2}$ $\ds \dfrac {\partial^2 u} {\partial y^2}$ $=$ $\ds \map {\dfrac \partial {\partial y} } {\dfrac 1 {x^2 + y} }$ $\ds$ $=$ $\ds -\dfrac 1 {\paren {x^2 + y}^2}$ $\ds \dfrac {\partial^3 u} {\partial x \partial y^2}$ $=$ $\ds \map {\dfrac \partial {\partial x} } {-\dfrac 1 {\paren {x^2 + y}^2} }$ $\ds$ $=$ $\ds -2 x \paren {-\dfrac 2 {\paren {x^2 + y}^3} }$ $\ds$ $=$ $\ds -\dfrac {4 x} {\paren {x^2 + y}^3}$ $\ds \dfrac {\partial^3 u} {\partial y \partial x \partial y}$ $=$ $\ds \map {\dfrac \partial {\partial y} } {-\dfrac {2 x} {\paren {x^2 + y}^2} }$ $\ds$ $=$ $\ds -2 x \paren {-\dfrac 2 {\paren {x^2 + y}^3} }$ $\ds$ $=$ $\ds -\dfrac {4 x} {\paren {x^2 + y}^3}$ $\ds \dfrac {\partial^3 u} {\partial y^2 \partial x}$ $=$ $\ds \map {\dfrac \partial {\partial y} } {-\dfrac {2 x} {\paren {x^2 + y}^2} }$ $\ds$ $=$ $\ds -2 x \paren {-\dfrac 2 {\paren {x^2 + y}^3} }$ $\ds$ $=$ $\ds -\dfrac {4 x} {\paren {x^2 + y}^3}$

$\blacksquare$