Partial Derivatives of u^2 + x^2 + y^2, u - v^3 + 3 x

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Theorem

Let:

\(\ds u^2 + x^2 + y^2\) \(=\) \(\ds 3\)
\(\ds u - v^3 + 3 x\) \(=\) \(\ds 4\)

Then:

\(\ds \dfrac {\partial u} {\partial x}\) \(=\) \(\ds -\dfrac x u\)
\(\ds \dfrac {\partial u} {\partial y}\) \(=\) \(\ds -\dfrac y u\)
\(\ds \dfrac {\partial v} {\partial x}\) \(=\) \(\ds \dfrac {x - 3 u} {3 u v^2}\)
\(\ds \dfrac {\partial v} {\partial y}\) \(=\) \(\ds \dfrac y {3 u v^2}\)


Proof

\(\ds 2 u \dfrac {\partial u} {\partial x} + 2 x\) \(=\) \(\ds 0\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial x}\) \(=\) \(\ds -\dfrac x u\)

$\Box$


\(\ds 2 u \dfrac {\partial u} {\partial y} + 2 y\) \(=\) \(\ds 0\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial y}\) \(=\) \(\ds -\dfrac y u\)

$\Box$


\(\ds \dfrac {\partial u} {\partial x} + 3 v^2 \dfrac {\partial v} {\partial x} + 3\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds -\dfrac x u + 3 v^2 \dfrac {\partial v} {\partial x} + 3\) \(=\) \(\ds 0\) substituting for $\dfrac {\partial u} {\partial x}$ from $(1)$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial v} {\partial x}\) \(=\) \(\ds \dfrac {x - 3 u} {3 u v^2}\) rearranging

$\Box$


\(\ds \dfrac {\partial u} {\partial y} + 3 v^2 \dfrac {\partial v} {\partial y}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds -\dfrac y u + 3 v^2 \dfrac {\partial v} {\partial x}\) \(=\) \(\ds 0\) substituting for $\dfrac {\partial u} {\partial y}$ from $(2)$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial v} {\partial y}\) \(=\) \(\ds \dfrac y {3 u v^2}\) rearranging

$\blacksquare$


Sources