Partial Derivative/Examples/u^2 + x^2 + y^2 = a^2

From ProofWiki
Jump to navigation Jump to search

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