Partial Derivatives of x^y^z
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Theorem
Let:
- $u = x^{\paren {y^z} }$
Then:
\(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds y^z x^{\paren {y^z - 1} }\) | ||||||||||||
\(\ds \dfrac {\partial u} {\partial y}\) | \(=\) | \(\ds x^{y^z} z y^{z - 1} \ln x\) | ||||||||||||
\(\ds \dfrac {\partial u} {\partial z}\) | \(=\) | \(\ds x^{\paren {y^z} } y^z \ln x \ln y\) |
Proof
\(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds y^z x^{\paren {y^z - 1} }\) | Power Rule for Derivatives, keeping $y$ and $z$ constant |
$\Box$
\(\ds u\) | \(=\) | \(\ds x^{\paren {y^z} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \ln u\) | \(=\) | \(\ds \map \ln {x^{\paren {y^z} } }\) | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds y^z \ln x\) | Logarithm of Power | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 u \dfrac {\partial u} {\partial y}\) | \(=\) | \(\ds z y^{z - 1} \ln x\) | Derivative of Natural Logarithm, Power Rule for Derivatives, Chain Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial u} {\partial y}\) | \(=\) | \(\ds u z y^{z - 1} \ln x\) | multiplying both sides by $u$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial u} {\partial y}\) | \(=\) | \(\ds x^{\paren {y^z} } z y^{z - 1} \ln x\) | substituting for $u$ |
$\Box$
\(\ds \ln u\) | \(=\) | \(\ds y^z \ln x\) | from $(1)$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 u \dfrac {\partial u} {\partial z}\) | \(=\) | \(\ds y^z \ln x \ln y\) | Derivative of Natural Logarithm, Derivative of General Exponential Function, Chain Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial u} {\partial z}\) | \(=\) | \(\ds u y^z \ln x \ln y\) | multiplying both sides by $u$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial u} {\partial z}\) | \(=\) | \(\ds x^{\paren {y^z} } y^z \ln x \ln y\) | substituting for $u$ |
$\blacksquare$
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $8$