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

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 \ \$ $\ds \dfrac {\partial u} {\partial y}$ $=$ $\ds -\dfrac y u$

$\blacksquare$