Theorem: Third Partial Derivatives of x^y

Let:
:$u = x^y$
Then:
:$\dfrac {\partial^3 u} {\partial x^2 \partial y} = \dfrac {\partial^3 u} {\partial x \partial y \partial x}$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial u} {\partial y}
      | r = x^y \ln x
      | c = Derivative of General Logarithm Function keeping $x$ constant
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\partial^2 u} {\partial x \partial y}
      | r = \map {\dfrac \partial {\partial x} } {\dfrac {\partial u} {\partial y} }
      | c = {{Defof|Second Partial Derivative}}
}}
{{eqn | r = y x^{y - 1} \ln x + \dfrac 1 x x^y
      | c = Power Rule for Derivatives, Derivative of Natural Logarithm keeping $y$ constant
}}
{{eqn | r = x^{y - 1} \paren {y \ln x + 1}
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\partial^3 u} {\partial x^2 \partial y}
      | r = \map {\dfrac \partial {\partial x} } {\dfrac {\partial^2 u} {\partial x \partial y} }
      | c = {{Defof|Third Partial Derivative}}
}}
{{eqn | r = x^{y - 1} \map {\dfrac \partial {\partial x} } {y \ln x + 1} + \paren {y \ln x + 1} \map {\dfrac \partial {\partial x} } {x^{y - 1} }
      | c = Product Rule for Derivatives
}}
{{eqn | r = x^{y - 1} \paren {\dfrac y x} + \paren {y \ln x + 1} \paren {y - 1} x^{y - 2}
      | c = Derivative of Natural Logarithm, Power Rule for Derivatives keeping $y$ constant
}}
{{eqn | r = x^{y - 2} \paren {y + \paren {y \ln x + 1} \paren {y - 1} }
      | c = simplifying
}}
{{eqn | r = x^{y - 2} \paren {y + y^2 \ln x + y -y \ln x - 1}
      | c = multiplying out
}}
{{eqn | r = x^{y - 2} \paren {2 y + y \paren {y - 1} \ln x - 1}
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = \dfrac {\partial u} {\partial x}
      | r = y x^{y - 1}
      | c = Power Rule for Derivatives keeping $y$ constant
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\partial^2 u} {\partial y \partial x}
      | r = \map {\dfrac \partial {\partial y} } {\dfrac {\partial u} {\partial x} }
      | c = {{Defof|Second Partial Derivative}}
}}
{{eqn | r = x^{y - 1} \map {\dfrac \partial {\partial y} } y + y \map {\dfrac \partial {\partial y} } {x^{y - 1} }
      | c = Product Rule for Derivatives
}}
{{eqn | r = x^{y - 1} + y x^{y - 1} \ln x
      | c = Derivative of Identity Function, Derivative of General Logarithm Function keeping $x$ constant
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\partial^3 u} {\partial x \partial y \partial x}
      | r = \map {\dfrac \partial {\partial x} } {\dfrac {\partial^2 u} {\partial y \partial x} }
      | c = {{Defof|Third Partial Derivative}}
}}
{{eqn | r = \map {\dfrac \partial {\partial x} } {x^{y - 1} + y x^{y - 1} \ln x}
      | c = 
}}
{{eqn | r = \paren {y - 1} x^{y - 2} + y \paren {x^{y - 1} \map {\dfrac \partial {\partial x} } {\ln x} + \ln x \map {\dfrac \partial {\partial x} } {x^{y - 1} } }
      | c = Power Rule for Derivatives, Product Rule for Derivatives
}}
{{eqn | r = \paren {y - 1} x^{y - 2} + y x^{y - 1} \paren {\dfrac 1 x} + y \ln x \paren {y - 1} x^{y - 2}
      | c = Derivative of Natural Logarithm, Power Rule for Derivatives
}}
{{eqn | r = x^{y - 2} \paren {\paren {y - 1} + y + y \ln x \paren {y - 1} }
      | c = factoring
}}
{{eqn | r = x^{y - 2} \paren {2 y + y \paren {y - 1} \ln x - 1}
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
The two expressions are equal.
Hence the result.
{{qed}}