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