hcju/proofwikiqa huggingface dataset:
Partial differentiation {{WRT|Differentiation}} $y$ gives: {{begin-eqn}} {{eqn | l = \dfrac {\partial u} {\partial y} - \dfrac {\partial v} {\partial y} + 2 \dfrac {\partial w} {\partial y}       | r = 0 }} {{eqn | l = 2 \dfrac {\partial u} {\partial y} + \dfrac {\partial v} {\partial y} - 2 \dfrac {\partial w} {\partial y}       | r = 0 }} {{eqn | l = \dfrac {\partial u} {\partial y} - \dfrac {\partial v} {\partial y} + \dfrac {\partial w} {\partial y}       | r = -1 }} {{end-eqn}} which can be expressed in matrix form as: :$\begin {pmatrix} 1 & -1 & 2 \\ 2 & 1 & -2 \\ 1 & -1 & 1 \end {pmatrix} \begin {pmatrix} \dfrac {\partial u} {\partial y} \\ \dfrac {\partial v} {\partial y} \\ \dfrac {\partial w} {\partial y} \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ -1 \end {pmatrix}$ Solving by Cramer's Rule: {{explain|Actually this is not Cramer's Rule, this will need to be taken care of}} {{begin-eqn}} {{eqn | o =        | r = \paren {\begin {array} {ccc {{|}} c} 1 & -1 & 2 & 0 \\ 2 & 1 & -2 & 0 \\ 1 & -1 & 1 & -1 \end {array} }       | c =  }} {{eqn | ll= \leadsto       | o =        | r = \paren {\begin {array} {ccc {{|}} c} 1 & -1 & 2 & 0 \\ 0 & 3 & -6 & 0 \\ 0 & 0 & -1 & -1 \end {array} }       | c = $\text r 2 \to \text r 2 - 2 \text r 1$, $\text r 3 \to \text r 3 - \text r 1$ }} {{eqn | ll= \leadsto       | o =        | r = \paren {\begin {array} {ccc {{|}} c} 1 & -1 & 2 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 1 \end {array} }       | c = $\text r 2 \to \text r 2 / 3$, $\text r 3 \to -\text r 3$ }} {{eqn | ll= \leadsto       | o =        | r = \paren {\begin {array} {ccc {{|}} c} 1 & 0 & 0 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 1 \end {array} }       | c = $\text r 1 \to \text r 1 - \text r 2$ }} {{eqn | ll= \leadsto       | o =        | r = \paren {\begin {array} {ccc {{|}} c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 1 \end {array} }       | c = $\text r 2 \to \text r 2 + 2 \text r 3$ }} {{end-eqn}} The solution can be read directly: {{begin-eqn}} {{eqn | l = \dfrac {\partial u} {\partial y}       | r = 0 }} {{eqn | l = \dfrac {\partial v} {\partial y}       | r = 2 }} {{eqn | l = \dfrac {\partial w} {\partial y}       | r = 1 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {f_1} {x, y}       | r = \map {\dfrac \partial {\partial x} } {x \map \arctan {x^2 + y} }       | c = {{Defof|Partial Derivative}} }} {{eqn | r = \map \arctan {x^2 + y} \map {\dfrac \partial {\partial x} } x + x \map {\dfrac \partial {\partial x} } {\map \arctan {x^2 + y} }       | c = Product Rule for Derivatives }} {{eqn | r = \map \arctan {x^2 + y} + x \cdot \dfrac 1 {1 + \paren {x^2 + y}^2} \cdot 2 x       | c = Product Rule for Derivatives, Derivative of Arctangent Function, Chain Rule for Derivatives }} {{eqn | r = \map \arctan {x^2 + y} + \dfrac {2 x^2} {1 + \paren {x^2 + y}^2}       | c = simplifying }} {{eqn | ll= \leadsto       | l = \map {f_1} {1, 0}       | r = \map \arctan {1^2 + 0} + \dfrac {2 \times 1^2} {1 + \paren {1^2 + 0}^2}       | c = substituting $x = 1$ and $y = 0$ }} {{eqn | r = \map \arctan 1 + \dfrac 2 2       | c = simplifying }} {{eqn | r = \dfrac \pi 4 + 1       | c = Tangent of $45 \degrees$ }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \map {f_2} {x, y}       | r = \map {\dfrac \partial {\partial y} } {x \map \arctan {x^2 + y} }       | c = {{Defof|Partial Derivative}} }} {{eqn | r = x \dfrac 1 {1 + \paren {x^2 + y}^2} \cdot 1       | c = Derivative of Identity Function, Derivative of Arctangent Function, Chain Rule for Derivatives }} {{eqn | r = \dfrac x {1 + \paren {x^2 + y}^2}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {f_1} {x, y, z}       | r = \ln y^2       | c = Derivative of Constant Multiple, Derivative of Identity Function, holding $y$ and $z$ constant }} {{eqn | ll= \leadsto       | l = \map {f_1} {1, -1, 0}       | r = \map \ln {\paren {-1}^2}       | c = substituting $\tuple {1, -1, 0}$ for $\tuple {x, y, z}$ }} {{eqn | r = \ln 1       | c = simplifying }} {{eqn | r = 0       | c = Natural Logarithm of 1 is 0 }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \map {f_2} {x, y, z}       | r = \map {\dfrac \partial {\partial y} } {x \ln y^2 + y e^z}       | c = {{Defof|Partial Derivative}} }} {{eqn | r = \dfrac x {y^2} \cdot 2 y + e^z       | c = Derivative of Natural Logarithm, Power Rule for Derivatives, Chain Rule for Derivatives, holding $x$ and $z$ constant }} {{eqn | r = \dfrac {2 x} y + e^z       | c = simplification }} {{eqn | ll= \leadsto       | l = \map {f_2} {x, x y, y + z}       | r = \dfrac {2 x} {x y} + e^{y + z}       | c = substituting $\tuple {x, x y, y + z}$ for $\tuple {x, y, z}$ }} {{eqn | r = \dfrac 2 y + e^{y + z}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {\partial u} {\partial x}       | r = y^z x^{\paren {y^z - 1} }       | c = Power Rule for Derivatives, keeping $y$ and $z$ constant }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = u       | r = x^{\paren {y^z} }       | c =  }} {{eqn | ll= \leadsto       | l = \ln u       | r = \map \ln {x^{\paren {y^z} } }       | c =  }} {{eqn | n = 1       | r = y^z \ln x       | c = Logarithm of Power }} {{eqn | ll= \leadsto       | l = \frac 1 u \dfrac {\partial u} {\partial y}       | r = z y^{z - 1} \ln x       | c = Derivative of Natural Logarithm, Power Rule for Derivatives, Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial y}       | r = u z y^{z - 1} \ln x       | c = multiplying both sides by $u$ }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial y}       | r = x^{\paren {y^z} } z y^{z - 1} \ln x       | c = substituting for $u$ }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \ln u       | r = y^z \ln x       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = \frac 1 u \dfrac {\partial u} {\partial z}       | r = y^z \ln x \ln y       | c = Derivative of Natural Logarithm, Derivative of General Exponential Function, Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial z}       | r = u y^z \ln x \ln y       | c = multiplying both sides by $u$ }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial z}       | r = x^{\paren {y^z} } y^z \ln x \ln y       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = u       | r = x^u + u^y       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = \ln u       | r = u \ln x + y \ln u       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 u \dfrac {\partial u} {\partial x}       | r = \frac u x + \ln x \dfrac {\partial u} {\partial x} + \frac y u \dfrac {\partial u} {\partial x}       | c = Derivative of Natural Logarithm, Chain Rule for Derivatives keeping $y$ constant }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial x} \paren {\frac 1 u - \ln x - \frac y u}       | r = \frac u x       | c = rearranging }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial x} \paren {\frac {1 - u \ln x - y} u}       | r = \frac u x       | c = rearranging }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial x}       | r = \frac {u^2} {x \paren {1 - u \ln x - y} }       | c = rearranging }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \ln u       | r = u \ln x + y \ln u       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = \frac 1 u \dfrac {\partial u} {\partial y}       | r = \ln x \dfrac {\partial u} {\partial y} + \ln u + \frac y u \dfrac {\partial u} {\partial y}       | c = Derivative of Natural Logarithm, Product Rule for Derivatives, Chain Rule for Derivatives keeping $x$ constant }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial y} \paren {\frac 1 u - \ln x - \frac y u}       | r = \ln u       | c = rearranging }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial y} \paren {\frac {1 - u \ln x - y} u}       | r = \ln u       | c = rearranging }} {{eqn | ll= \leadsto       | l = \dfrac {\partial u} {\partial y}       | r = \frac {u \ln u} {1 - u \ln x - y}       | c = rearranging }} {{end-eqn}} {{qed}}
{{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}}