Partial Derivatives of x^u + u^y

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Theorem

Let:

$u = x^u + u^y$

Then:

\(\ds \dfrac {\partial u} {\partial x}\) \(=\) \(\ds \frac {u^2} {x \paren {1 - u \ln x - y} }\)
\(\ds \dfrac {\partial u} {\partial y}\) \(=\) \(\ds \frac {u \ln u} {1 - u \ln x - y}\)


Proof

\(\ds u\) \(=\) \(\ds x^u + u^y\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \ln u\) \(=\) \(\ds u \ln x + y \ln u\)
\(\ds \leadsto \ \ \) \(\ds \frac 1 u \dfrac {\partial u} {\partial x}\) \(=\) \(\ds \frac u x + \ln x \dfrac {\partial u} {\partial x} + \frac y u \dfrac {\partial u} {\partial x}\) Derivative of Natural Logarithm, Chain Rule for Derivatives keeping $y$ constant
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial x} \paren {\frac 1 u - \ln x - \frac y u}\) \(=\) \(\ds \frac u x\) rearranging
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial x} \paren {\frac {1 - u \ln x - y} u}\) \(=\) \(\ds \frac u x\) rearranging
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial x}\) \(=\) \(\ds \frac {u^2} {x \paren {1 - u \ln x - y} }\) rearranging

$\Box$


\(\ds \ln u\) \(=\) \(\ds u \ln x + y \ln u\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds \frac 1 u \dfrac {\partial u} {\partial y}\) \(=\) \(\ds \ln x \dfrac {\partial u} {\partial y} + \ln u + \frac y u \dfrac {\partial u} {\partial y}\) Derivative of Natural Logarithm, Product Rule for Derivatives, Chain Rule for Derivatives keeping $x$ constant
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial y} \paren {\frac 1 u - \ln x - \frac y u}\) \(=\) \(\ds \ln u\) rearranging
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial y} \paren {\frac {1 - u \ln x - y} u}\) \(=\) \(\ds \ln u\) rearranging
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial y}\) \(=\) \(\ds \frac {u \ln u} {1 - u \ln x - y}\) rearranging

$\blacksquare$


Sources