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
| \(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds \dfrac 1 {x^2 + y} \cdot 2 x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {2 x} {x^2 + y}\) | ||||||||||||
| \(\ds \dfrac {\partial u} {\partial y}\) | \(=\) | \(\ds \dfrac 1 {x^2 + y} \cdot 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {x^2 + y}\) | ||||||||||||
| \(\ds \dfrac {\partial^2 u} {\partial y \partial x}\) | \(=\) | \(\ds \map {\dfrac \partial {\partial y} } {\dfrac {2 x} {x^2 + y} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\dfrac {2 x} {\paren {x^2 + y}^2}\) | ||||||||||||
| \(\ds \dfrac {\partial^2 u} {\partial x \partial y}\) | \(=\) | \(\ds \map {\dfrac \partial {\partial x} } {\dfrac 1 {x^2 + y} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\dfrac {2 x} {\paren {x^2 + y}^2}\) | ||||||||||||
| \(\ds \dfrac {\partial^2 u} {\partial y^2}\) | \(=\) | \(\ds \map {\dfrac \partial {\partial y} } {\dfrac 1 {x^2 + y} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\dfrac 1 {\paren {x^2 + y}^2}\) | ||||||||||||
| \(\ds \dfrac {\partial^3 u} {\partial x \partial y^2}\) | \(=\) | \(\ds \map {\dfrac \partial {\partial x} } {-\dfrac 1 {\paren {x^2 + y}^2} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -2 x \paren {-\dfrac 2 {\paren {x^2 + y}^3} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\dfrac {4 x} {\paren {x^2 + y}^3}\) | ||||||||||||
| \(\ds \dfrac {\partial^3 u} {\partial y \partial x \partial y}\) | \(=\) | \(\ds \map {\dfrac \partial {\partial y} } {-\dfrac {2 x} {\paren {x^2 + y}^2} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -2 x \paren {-\dfrac 2 {\paren {x^2 + y}^3} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\dfrac {4 x} {\paren {x^2 + y}^3}\) | ||||||||||||
| \(\ds \dfrac {\partial^3 u} {\partial y^2 \partial x}\) | \(=\) | \(\ds \map {\dfrac \partial {\partial y} } {-\dfrac {2 x} {\paren {x^2 + y}^2} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -2 x \paren {-\dfrac 2 {\paren {x^2 + y}^3} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\dfrac {4 x} {\paren {x^2 + y}^3}\) |
$\blacksquare$
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: $1.3$ Higher Order Derivatives: Example $\text D$