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
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $9$