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
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $10$