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:

\(\displaystyle \dfrac {\partial u} {\partial x}\) \(=\) \(\displaystyle \dfrac {u y} {u + 1}\)
\(\displaystyle \dfrac {\partial u} {\partial y}\) \(=\) \(\displaystyle \dfrac {u x} {u + 1}\)


Proof

\(\displaystyle u + \ln u\) \(=\) \(\displaystyle x y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac \partial {\partial x} u + \dfrac \partial {\partial x} \ln u\) \(=\) \(\displaystyle \dfrac \partial {\partial x} x y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\partial u} {\partial x} + \dfrac 1 u \dfrac {\partial u} {\partial x}\) \(=\) \(\displaystyle y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\partial u} {\partial x} \paren {1 + \dfrac 1 u}\) \(=\) \(\displaystyle y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\partial u} {\partial x}\) \(=\) \(\displaystyle \dfrac {u y} {u + 1}\)

and:

\(\displaystyle u + \ln u\) \(=\) \(\displaystyle x y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac \partial {\partial y} u + \dfrac \partial {\partial y} \ln u\) \(=\) \(\displaystyle \dfrac \partial {\partial y} x y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\partial u} {\partial y} + \dfrac 1 u \dfrac {\partial u} {\partial y}\) \(=\) \(\displaystyle x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\partial u} {\partial y} \paren {1 + \dfrac 1 u}\) \(=\) \(\displaystyle x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\partial u} {\partial y}\) \(=\) \(\displaystyle \dfrac {u x} {u + 1}\)

$\blacksquare$


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