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