hcju/proofwikiqa huggingface dataset:
For $\struct {\Z_{\ge 0}, +, \times}$ to be a ring, it is necessary for the algebraic structure $\struct {\Z_{\ge 0}, +}$ to form a group. But from the corollary to Natural Numbers under Addition do not form Group: :$\struct {\Z_{\ge 0}, +}$ is not a group. {{qed}}
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup. For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed. That is: :$\forall x, y \in S: x \times y \in S$ Let $x = \dfrac 1 2$ and $y = \dfrac 3 2$. Both $x$ and $y$ are in $S$, as both are rational numbers expressed in canonical form whose denominators are not divisible by $4$. But then: :$x \times y = \dfrac 1 2 \times \dfrac 3 2 = \dfrac 3 4$ which is a rational numbers expressed in canonical form whose denominator ''is'' divisible by $4$. Hence $x \times y \notin S$ and so $\struct {S, \times}$ is not closed. The result follows. {{qed}}
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup. For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed. That is: :$\forall x, y \in S: x \times y \in S$ Let $x = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix}$ and $y = \begin {pmatrix} 0 & 2 \\ 2 & 0 \end {pmatrix}$. Both $x$ and $y$ are in $S$, as both are square matrices of order $2$ whose diagonal elements are zero. But then: :$x y = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix} \begin {pmatrix} 0 & 2 \\ 2 & 0 \end {pmatrix} = \begin {pmatrix} 2 & 0 \\ 0 & 2 \end {pmatrix}$ which is a square matrix of order $2$ whose diagonal elements are ''not'' zero. Hence $x y \notin S$ and so $\struct {S, \times}$ is not closed. The result follows. {{qed}}
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup. For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed. That is: :$\forall x, y \in S: x \times y \in S$ Let $\mathbf A$ and $\mathbf B$ be elements of $S$. The matrix multiplication operation is defined on $\mathbf A$ and $\mathbf B$ {{iff}} $\mathbf A$ is of order $m \times p$ and $\mathbf A$ is of order $p \times n$, for some $m, n, p \in \N_{>0}$. That is, the second dimension of $\mathbf A$ must be equal to the first dimension of $\mathbf B$. But in this case, the second dimension of $\mathbf A$ is $n$, and the first dimension of $\mathbf B$ is $m$. As we have been given that $m \ne n$, it follows that matrix multiplication operation is not defined on $S$. Hence the result. {{qed}}
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup. For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\times$ is associative on $S$. However, from Vector Cross Product is not Associative, this is not the case here. The result follows. {{qed}}
It is to be demonstrated that $\struct {\Z, -}$ does not satisfy the semigroup axioms. We then have Subtraction on Numbers is Not Associative. So, for example: :$3 - \paren {2 - 1} = 2 \ne \paren {3 - 2} - 1 = 0$ Thus it has been demonstrated that $\struct {\Z, -}$ does not satisfy {{SemigroupAxiom|1}}. Hence the result. {{qed}}
===== Lemma ===== {{:Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Sufficient Condition/Lemma}}{{qed|lemma}}  Let: :$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$ It is to be shown that: :* $\quad \mathscr I$ satisfies the matroid axioms and :* $\rho$ is the rank function of the matroid $M = \struct{S, \mathscr I}$ === Matroid Axioms === ==== Matroid Axiom $(\text I 1)$ ==== From Cardinality of Empty Set: :$\card \O = 0$ By rank axiom $(\text R 1)$: :$\map \rho \O = 0$ Hence: :$\map \rho \O = \card \O$ So: :$\O \in \mathscr I$ Hence: $M$ satisfies matroid axiom $(\text I 1)$. {{qed|lemma}} ==== Matroid Axiom $(\text I 2)$ ==== Let :$X \in \mathscr I$ {{AimForCont}} :$\exists Y \subseteq X : Y \notin \mathscr I$ Let: :$Y_0 \subseteq X : \card {Y_0} = \max \set{\card Z : Z \subseteq X \land Z \notin \mathscr I}$ From lemma: :$\map \rho {Y_0} < \card {Y_0}$ As $\map \rho X = \card X$: :$Y_0 \ne X$ From Set Difference with Proper Subset is Proper Subset: :$X \setminus Y_0 \neq \O$ Let $y \in X \setminus Y_0$. We have: {{begin-eqn}} {{eqn | l = \card {Y_0} + 1       | r = \card {Y_0 \cup \set y}       | c = Corollary of Cardinality of Set Union }} {{eqn | r = \map \rho {Y_0 \cup \set y}       | c = As $\card {Y_0} = \max \set{\card Z : Z \subseteq X \land Z \notin \mathscr I}$ }} {{eqn | o = \le       | r = \map \rho {Y_0} + 1       | c = Rank axiom $(\text R 2)$ }} {{eqn | o = <       | r = \card {Y_0} + 1       | c = As $\map \rho {Y_0} < \card {Y_0}$ }} {{end-eqn}} This is a contradiction. So: :$\forall Y \subseteq X : Y \in \mathscr I$ Hence: :$M$ satisfies matroid axiom $(\text I 2)$. {{qed|lemma}} ==== Matroid Axiom $(\text I 3)$ ==== Let :$U \in \mathscr I$ :$V \subseteq S$ :$\card U < \card V$ We prove the contrapositive statement: :$\paren{\forall x \in V \setminus U : U \cup \set x \notin \mathscr I} \implies V \notin \mathscr I$ Let: :$\forall x \in V \setminus U : U \cup \set x \notin \mathscr I$ {{qed|lemma}} === $\rho$ is Rank Function ===
From Independent Subset is Contained in Maximal Independent Subset: :$\exists Y \in \mathscr I : X \subseteq Y \subseteq A : \card Y = \map \rho A$ By definition of a maximal independent Subset of $A$: :$X = Y$ The result follows. {{qed}} Category:Matroid Theory 4332psornx28hd65ro444fgft84ne0w
{{begin-eqn}} {{eqn | l = \mathbf F       | r = \map {\dfrac \d {\d t} } {m \mathbf v}       | c = Newton's Second Law of Motion }} {{eqn | r = m \dfrac {\d \mathbf v} {\d t} + \mathbf v \dfrac {\d m} {\d t}       | c =  }} {{eqn | r = m \dfrac {\d \mathbf v} {\d t} + 0       | c = Derivative of Constant }} {{eqn | r = m \mathbf a       | c = {{Defof|Acceleration}} }} {{end-eqn}} {{qed}}
By the definition of like vector quantities: :$\mathbf a$ and $\mathbf b$ are '''like vector quantities''' {{iff}} they have the same direction. By definition of unit vector: :$\dfrac {\mathbf a} {\size {\mathbf a} } = \dfrac {\mathbf b} {\size {\mathbf b} }$ as both are in the same direction, and both have length $1$. By definition of scalar division: :$\dfrac 1 {\size {\mathbf a} } \mathbf a = \dfrac 1 {\size {\mathbf b} } \mathbf b$ Hence, multiplying by $\size {\mathbf a}$: :$\mathbf a = \dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$ {{qed}}
=== Positive Definiteness === Let $\tuple {x , y} \in V$. Then: {{begin-eqn}} {{eqn | l = \norm {\tuple {x, y} }        | r = \map \max {\norm x, \norm y}       | c = {{defof|Direct Product Norm}} }} {{eqn | o = \ge       | r =  0       | c = Norm Axiom $N1$: Positive Definiteness }} {{end-eqn}} Suppose $\norm {\tuple {x, y} } = 0$. Then: {{begin-eqn}} {{eqn | l = 0        | o = \le       | r = \norm x        | c = Norm Axiom $N1$: Positive Definiteness       }} {{eqn | o = \le        | r = \map \max {\norm x, \norm y}  }} {{eqn | r = \norm {\tuple {x, y} }       | c = {{defof|Direct Product Norm}} }} {{eqn | r = 0 }} {{end-eqn}} Hence, $\norm x = 0$ {{qed|lemma}} === Positive Homogeneity === {{begin-eqn}} {{eqn | l = \norm {\alpha \tuple {x, y} }       | r = \norm {\tuple {\alpha x, \alpha y} }       | c = Induced component-wise operations }} {{eqn | r = \map \max {\norm {\alpha x}, \norm {\alpha y} }        | c = {{defof|Direct Product Norm}} }} {{eqn | r = \map \max {\size \alpha \norm x, \size \alpha \norm y}        | c = Norm Axiom $N2$ : Positive Homogeneity }} {{eqn | r = \size \alpha \max \set {\norm x, \norm y}  }} {{eqn | r = \size \alpha \norm {\tuple {x ,y} }       | c = {{defof|Direct Product Norm}} }} {{end-eqn}} {{qed|lemma}} === Triangle Inequality === {{begin-eqn}} {{eqn | l = \norm {x_1 + x_2}        | o = \le        | r = \norm {x_1} + \norm {x_2}  }} {{eqn | o = \le       | r = \map \max {\norm {x_1}, \norm {y_1} } + \map \max {\norm {x_2}, \norm {y_2} }  }} {{eqn | r = \norm {\tuple {x_1, y_1} }+ \norm {\tuple {x_2, y_2} }       | c = {{defof|Direct Product Norm}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \norm {y_1 + y_2}        | o = \le        | r = \norm {y_1} + \norm {y_2}  }} {{eqn | o = \le        | r = \map \max {\norm {x_1}, \norm {y_1} } + \map \max {\norm {x_2}, \norm {y_2} }  }} {{eqn | r = \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} }       | c = {{defof|Direct Product Norm}} }} {{end-eqn}} Together: :$\map \max {\norm {x_1 + x_2}, \norm {y_1 + y_2}} \le \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} }$ Then: {{begin-eqn}} {{eqn | l = \norm {\tuple {x_1, y_1} + \tuple {x_2, y_2} }        | r = \norm {\tuple {x_1 + x_2, y_1 + y_2} }        | c = Induced component-wise operations }} {{eqn | r = \map \max {\norm {x_1 + x_2}, \norm {y_1 + y_2} }        | c = {{defof|Direct Product Norm}} }} {{eqn | o = \le        | r = \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} } }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf a       | r = \size {\mathbf a} \hat {\mathbf a}       | c = Vector Quantity as Scalar Product of Unit Vector Quantity }} {{eqn | ll= \leadsto       | l = m \mathbf a       | r = m \paren {\size {\mathbf a} \hat {\mathbf a} }       | c =  }} {{end-eqn}} Then: {{finish|hard to prove something trivial}} {{qed}}
By definition of like vector quantities, $\mathbf a$ and $\mathbf b$ have the same direction. By definition of unit vector, $\mathbf {\hat a}$ and $\mathbf {\hat b}$ are both of magnitude $1$. Hence the result, by Equality of Vector Quantities. {{qed}} Category:Vectors 4dc6u0f04r1xrbj90fu5p7eaj30d0uk
400px Let $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ be unit vectors in the directions of $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively. Let $O$ be a point in space. Take $\vec {OP} := \mathbf r$. With $OP$ as its space diagonal, construct a parallelepiped with edges $OA$, $OB$ and $OC$ parallel to $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ respectively. Only one such parallelepiped can be so constructed. Let $x$, $y$ and $z$ be the length of the edges $OA$, $OB$ and $OC$ respectively. Then: {{begin-eqn}} {{eqn | l = \mathbf r       | r = \vec {OA} + \vec {AF} + \vec {FP}       | c =  }} {{eqn | r = \vec {OA} + \vec {OB} + \vec {OC}       | c =  }} {{eqn | r = x \mathbf {\hat a} + y \mathbf {\hat b} + z \mathbf {\hat c}       | c =  }} {{end-eqn}} Thus $\mathbf r$ is the resultant of the $3$ vector quantities $x \mathbf {\hat a}$, $y \mathbf {\hat b}$ and $z \mathbf {\hat c}$ which, by construction, are parallel to $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively. The fact that only one parallelepiped can be constructed in the above proves uniqueness. {{qed}}
Let $\mathbf a$ and $\mathbf b$ be vector quantities. Then by Vector Quantity can be Expressed as Sum of 3 Non-Coplanar Vectors, $\mathbf a$ and $\mathbf b$ can be expressed uniquely as components. So if $\mathbf a$ and $\mathbf b$ then the components of $\mathbf a$ are the same as the components of $\mathbf b$ Suppose $\mathbf a$ and $\mathbf b$ have the same components: $\mathbf x$, $\mathbf y$ and $\mathbf z$. Then by definition: :$\mathbf a = \mathbf x + \mathbf y + \mathbf z$ and also: :$\mathbf b = \mathbf x + \mathbf y + \mathbf z$ and trivially: :$\mathbf a = \mathbf b$ {{qed}}
=== Necessary Condition === Let $x \perp_B y$. Let $V' \subset V$ be the subspace spanned by $x$ and $y$. Define $\overline f$ on $V'$ as: :$\map {\overline f} {a x + b y} = a \norm x$ for $a$ and $b$ scalars. Clearly, $\overline f$ is linear and: :$\map {\overline f} x = \norm x$ :$\map {\overline f} y = 0$ Further: {{begin-eqn}} {{eqn | l = \norm {a x + b y}       | r = \size a \norm {x + \dfrac b a y}       | c =  }} {{eqn | o = \ge       | r = \size a \norm x       | c =  }} {{eqn | r = \size {\map {\overline f} {a x + b y} }       | c =  }} {{end-eqn}} proving that $\overline f$ is a bounded functional of norm $1$. Now by Hahn-Banach Theorem, $\overline f$ can be extended to a functional $f$ on $V$ such that $\norm f = \norm {\overline f} = 1$ This proves the necessity. {{qed|lemma}} === Sufficient Condition === Let such a functional $f$ on $V$ exist. Then for any scalar $\lambda$: :$\norm {x + \lambda y} \ge \size {\map f {x + \lambda y} } = \norm x$ establishing the sufficiency. {{qed}}
:480px By definition, the direction cosines are the cosines of the angles that $\mathbf r$ makes with the coordinate axes. By definition of the components of $\mathbf r$: :$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$ Thus: :$\mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$ and the result follows. {{qed}}
Let the initial point of $\mathbf r$ be $\tuple {x_1, y_1, z_1}$. Let the terminal point of $\mathbf r$ be $\tuple {x_2, y_2, z_2}$. Thus, by definition of the components of $\mathbf r$, the magnitude of $\mathbf r$ equals the distance between $\tuple {x_1, y_1, z_1}$ and $\tuple {x_2, y_2, z_2}$. The result follows from Distance Formula in 3 Dimensions. {{qed}}
By definition of the zero vector, the magnitude of $\mathbf r$ is equal to zero. By Magnitude of Vector Quantity in terms of Components: :$\size {\mathbf r} = \sqrt {x^2 + y^2 + z^2} = 0$ where $\size {\mathbf r}$ denotes the magnitude of $\mathbf r$. As each of $x$, $y$ and $z$ are real numbers, each of $x^2$, $y^2$ and $z^2$ is non-negative. so in order for $\sqrt {x^2 + y^2 + z^2} = 0$, it must follow that each of $x$, $y$ and $z$ is zero. Hence the result. {{qed}}
From Components of Vector in terms of Direction Cosines: :$(1): \quad \mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$ where $r$ denotes the magnitude of $\mathbf r$, that is: :$r := \size {\mathbf r}$ By definition of unit vector: :$\mathbf {\hat r} = \dfrac {\mathbf r} r$ The result follows by multiplication of both sides of $(1)$ by $\dfrac 1 r$. {{qed}}
Let $\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ be a Cauchy sequence in $V$: : $\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n > N: \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} } < \epsilon$ We have that: {{begin-eqn}} {{eqn | l = \norm {x_n - x_m}        | o = \le       | r = \map \max {\norm {x_n - x_m}, \norm {y_n - y_m} } }} {{eqn | r = \norm {\tuple {x_n - x_m, y_n - y_m} }       | c = {{defof|Direct Product Norm}} }} {{eqn | r =  \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} }       | c = Induced component-wise operations }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} Hence, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy sequence in $X$. By assumption, $X$ is a Banach space. Therefore, $\sequence {x_n}_{n \mathop \in \N}$ converges to $x \in X$. By analogous arguments, $\sequence {y_n}_{n \mathop \in \N}$ converges to $y \in Y$. Then there exists $\epsilon \in \R_{> 0}$ such that: :$\exists N_x \in \N : \forall n \in \N : n > N_x : \norm {x - x_n}_n < \epsilon$ :$\exists N_y \in \N : \forall n \in \N : n > N_y : \norm {y - y_n}_n < \epsilon$ Let $N := \map \max {N_x, N_y}$. Then: :$\forall n \in \N : n > N : \paren {\norm {x - x_n}_n < \epsilon} \land \paren{\norm {y - y_n}_n < \epsilon}$ Thus: {{begin-eqn}} {{eqn | l = \norm {\tuple {x, y} - \tuple {x_n, y_n} }        | r = \norm {\tuple {x - x_n, y - y_n} }  }} {{eqn | r = \map \max {\norm {x - x_n}, \norm {y - y_n} }        | c = {{defof|Direct Product Norm}} }} {{eqn | o = <        | r = \epsilon }} {{end-eqn}} Therefore, $\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ converges to $\tuple {x, y}$. By definition, $V$ is a Banach space. {{qed}}
:400px Let $\mathbf r$ be represented by a directed line segment whose initial point coincides with the origin $O$. Let the terminal point of $\mathbf r$ be identified with the point $P$. Let $\CC$ be rotated to $\CC'$ through an angle $\varphi$ as shown, keeping $P$ fixed. We have that: {{begin-eqn}} {{eqn | l = x'       | r = OA + BP       | c = by inspection }} {{eqn | r = x \cos \varphi + y \sin \varphi       | c = {{Defof|Cosine}}, {{Defof|Sine}} }} {{eqn | l = y'       | r = xB - xA       | c = by inspection }} {{eqn | r = y \cos \varphi - x \sin \varphi       | c = {{Defof|Cosine}}, {{Defof|Sine}} }} {{end-eqn}} hence the result. {{qed}}
{{AimForCont}}: :$\exists Y \subseteq S : \map \rho Y > \card Y$ Let: :$Y_0 \subseteq S : \card{Y_0} = \min \set{\card Y : \map \rho Y > \card Y}$ By rank axiom $(\text R 1)$: :$\map \rho \O = 0$ From Cardinality of Empty Set: :$\card \O = 0$ So: :$\map \rho \O = \card \O$ Hence: :$Y_0 \neq \O$ Let $y \in Y_0$. Let $Y = Y_0 \setminus \set y$. From Cardinality of Set Difference: :$\card Y = \card{Y_0} - 1$ By choice of $Y_0$: :$\map \rho Y \le \card Y$ By rank axiom $(\text R 2)$: :$\map \rho {Y_0} = \map \rho {Y \cup \set y} \le \map \rho Y + 1 \le \card Y + 1 = \card{Y_0}$ This contradicts the choice of $Y_0$: :$\card{Y_0} = \min \set{\card Y : \map \rho Y > \card Y}$ It follows that: :$\forall Y \subseteq S: \map \rho Y \le \card Y$ {{qed}} Category:Matroid Theory gd78tscazxtxbfdp6j5pb8sm0b03ofd
By definition of dot product: :$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \cos \theta$ where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$. When $\mathbf a$ and $\mathbf b$ be perpendicular, by definition $\theta = 90 \degrees$. The result follows by Cosine of Right Angle, which gives that $\cos 90 \degrees = 0$. {{qed}}
By definition of orthonormal basis: :$(1): \quad \tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is an orthogonal basis of $V$ :$(2): \quad \norm {\mathbf e_1} = \norm {\mathbf e_2} = \cdots = \norm {\mathbf e_1} = 1$ From $(1)$ we have by definition that $\mathbf e_i$ and $\mathbf e_j$ are perpendicular whenever $i \ne j$ Thus, from Dot Product of Perpendicular Vectors, for all $i$ and $j$, when $i \ne j$: :$\mathbf e_i \cdot \mathbf e_j = 0$ From Dot Product of Vector with Itself, for all $i$ we have: :$\mathbf e_i \cdot \mathbf e_i = \norm {\mathbf e_i}^2$ From $(2)$, we have that: :$\norm {\mathbf e_i} = 1$ and so: :$\mathbf e_i \cdot \mathbf e_i = 1$ Putting this together, we have: :$\forall i, j \in \set {1, 2, \ldots, n}: \mathbf e_i \cdot \mathbf e_j = \begin {cases} 1 & : i = j \\ 0 & : i \ne j \end {cases}$ The result follows by definition of the Kronecker delta. {{qed}}
From Closed Real Interval is Closed Set, $I$ is a closed set of $\R$. From Real Interval is Bounded in Real Numbers, $I$ is bounded in $\R$. The result follows by definition of compact. {{qed}}
We have that $\struct {\R, \size {\, \cdot \,}}$ is a normed vector space. Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence in $I$. $\sequence {a_n}$ is bounded by $a$ and $b$. By Bolzano-Weierstrass theorem, there exists a convergent subsequence $\sequence {a_{n_k}}_{k \in \N}$ with a limit $L$. Then: :$\forall k \in \N : a \le a_{n_k} \le b$ Take the limit $k \to \infty$: :$a \le L \le b$ Hence, $L \in I$. The result follows by definition of compact. {{qed}}
{{begin-eqn}} {{eqn | l = \frac \partial {\partial x} \frac {\sin x y} {\map \cos {x + y} }       | r = \frac {\map \cos {x + y} y \cos x y - \sin x y \paren {-\map \sin {x + y} } } {\map {\cos^2} {x + y} }       | c = Quotient Rule for Derivatives, treating $y$ as a constant }} {{eqn | r = \dfrac {y \map \cos {x + y} \cos x y + \map \sin {x + y} \sin x y} {\map {\cos^2} {x + y} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {\partial^2} {\partial r^2} \map \ln {r^2 + s}       | r = \map {\dfrac \partial {\partial r} } {\dfrac \partial {\partial r} \map \ln {r^2 + s} }       | c = {{Defof|Second Partial Derivative}} }} {{eqn | r = \dfrac \partial {\partial r} \frac 1 {r^2 + s} 2 r       | c = Derivative of Natural Logarithm, Chain Rule for Derivatives, treating $s$ as a constant }} {{eqn | r = \frac {2 \paren {r^2 + s} - 2 r \paren {2 r} } {\paren {r^2 + s}^2}       | c = Quotient Rule for Derivatives, again treating $s$ as a constant }} {{eqn | r = \frac {2 \paren {s - r^2} } {\paren {r^2 + s}^2}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {f_1} {x, y}       | r = \dfrac \partial {\partial x} {\map {\tan^2} {x^2 - y^2} }       | c = {{Defof|Partial Derivative}} }} {{eqn | r = 2 \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2} \cdot 2 x       | c = Derivative of Square of Tangent, Derivative of Square Function, Chain Rule for Derivatives }} {{eqn | r = 4 x \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2}       | c = simplifying }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \map {f_2} {x, y}       | r = \dfrac \partial {\partial y} {\map {\tan^2} {x^2 - y^2} }       | c = {{Defof|Partial Derivative}} }} {{eqn | r = 2 \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2} \cdot \paren {-2 y}       | c = Derivative of Square of Tangent, Derivative of Square Function, Chain Rule for Derivatives }} {{eqn | r = -4 y \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2}       | c = simplifying }} {{eqn | ll= \leadsto       | l = \map {f_2} {1, 2}       | r = -4 \paren 2 \map \tan {1^2 - 2^2} \map {\sec^2} {1^2 - 2^2}       | c =  }} {{eqn | r = -8 \map \tan {-3} \map {\sec^2} {-3}       | c =  }} {{eqn | r = -8 \paren {-\tan 3} \sec^2 3       | c = Tangent Function is Odd, Secant Function is Even }} {{eqn | r = 8 \tan 3 \sec^2 3       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\tan^2 x}       | r = 2 \tan x \map {\dfrac \d {\d x} } {\tan x}       | c = Chain Rule for Derivatives, Derivative of Square Function }} {{eqn | r = 2 \tan x \sec^2 x        | c = Derivative of Tangent Function }} {{end-eqn}} {{qed}} Category:Tangent Function nr7j1qx0qa2pkgzchptcsv4cp87ohvw
First we make sure that $\tuple {1, -2, -2}$ actually satisfies the equation: {{begin-eqn}} {{eqn | l = x^2 - 2 x y - 1       | r = \paren 1^2 - 2 \paren 1 \paren {-2} - 1       | c = at $\tuple {1, -2, -2}$ }} {{eqn | r = 1 + 4 - 1       | c =  }} {{eqn | r = 4       | c =  }} {{eqn | r = \paren {-2}^2       | c =  }} {{eqn | r = z^2       | c = at $\tuple {1, -2, -2}$ }} {{end-eqn}} and it is seen that this is the case. {{qed|lemma}} === Explicit Method === {{:Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Explicit Method}} === Implicit Method === {{:Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Implicit Method}} Category:Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2) Category:Examples of Partial Derivatives krx6ob9fe8q70qq2cpcss95dyambcgy
{{begin-eqn}} {{eqn | l = z^2       | r = x^2 - 2 x y - 1       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = \pm \sqrt {x^2 - 2 x y - 1}       | c =  }} {{end-eqn}} When $z = -2$ we have that $\sqrt {x^2 - 2 x y - 1} < 0$. Hence: :$z = -\sqrt {x^2 - 2 x y - 1}$ Thus we continue: {{begin-eqn}} {{eqn | ll= \leadsto       | l = \dfrac {\partial z} {\partial x}       | r = -\frac 1 {2 \sqrt {x^2 - 2 x y - 1} } \cdot \paren {2 x - 2 y}       | c = Power Rule for Derivatives, Chain Rule for Derivatives }} {{eqn | r = -\frac {2 x - 2 y} {2 \sqrt {x^2 - 2 x y - 1} }       | c =  }} {{eqn | r = -\frac {2 \paren 1 - 2 \paren {-2} } {2 \sqrt {\paren 1^2 - 2 \paren 1 \paren {-2} - 1} }       | c = substituting $x = 1$ and $y = -2$ }} {{eqn | r = -\frac 6 {2 \sqrt 4}       | c = simplifying }} {{eqn | r = -\frac 3 2       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z^2       | r = x^2 - 2 x y - 1       | c =  }} {{eqn | l = 2 z \dfrac {\partial z} {\partial x}       | r = 2 x - 2 y       | c = Power Rule for Derivatives, Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \dfrac {\partial z} {\partial x}       | r = \frac {2 \paren 1 - 2 \paren {-2} } {2 \paren {-2} }       | c = substituting $x = 1$, $y = -2$ and $z = -2$ }} {{eqn | r = \frac 6 {-4}       | c = simplifying }} {{eqn | r = -\frac 3 2       | c = simplifying }} {{end-eqn}} {{qed}}