Partial Derivative/Examples/u^2 + x^2 + y^2 = a^2
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Example of Partial Derivative
Let $u^2 + x^2 + y^2 = a^2$ be an implicit function.
Then:
\(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds -\dfrac x u\) | ||||||||||||
\(\ds \dfrac {\partial u} {\partial y}\) | \(=\) | \(\ds -\dfrac y u\) |
Proof
\(\ds u^2 + x^2 + y^2\) | \(=\) | \(\ds a^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac \partial {\partial x} u^2 + \dfrac \partial {\partial x} x^2 + \dfrac \partial {\partial x} y^2\) | \(=\) | \(\ds \dfrac \partial {\partial x} a^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 u \dfrac {\partial u} {\partial x} + 2 x + 0\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds -\dfrac x u\) |
and:
\(\ds u^2 + x^2 + y^2\) | \(=\) | \(\ds a^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac \partial {\partial y} u^2 + \dfrac \partial {\partial y} x^2 + \dfrac \partial {\partial y} y^2\) | \(=\) | \(\ds \dfrac \partial {\partial y} a^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 u \dfrac {\partial u} {\partial y} + 0 + 2 y\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial u} {\partial y}\) | \(=\) | \(\ds -\dfrac y u\) |
$\blacksquare$
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: $1.2$ Implicit Functions