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The number of e-folds $N=\log a$ satisfiesWhen $\varphi$ is monotone, the chain rule gives $dt=d\varphi /\dot{\varphi }$, and henceUsing the two slow-roll equations, $3H\dot{\varphi }\simeq -V^{\prime }(\varphi )$ and $H^2\simeq 8\pi GV/3$, eliminates both $H$ and time:This is the differential form of the slow-roll e-fold count.

By the multivariable chain rule and Hamilton's equations,The sum is the Poisson bracket $\{f,H\}$, giving

WriteBy the chain rule, ${\tilde{V}}^{\alpha }=J^{\alpha }{}_{\mu }{V}^{\mu }$, ${\tilde{W}}^{\alpha }=J^{\alpha }{}_{\mu }{W}^{\mu }$, and ${\tilde{V}}^{\beta }{\tilde{\partial }}_{\beta }=V^{\nu }{\partial }_{\nu }$. ThereforeThe last term vanishes because its first factor is antisymmetric in $\mu ,\nu$ and its second is symmetric. Thus the coordinate invariance of the Lie bracket is

For plane polar coordinates, $x=r\cos \theta$ and $y=r\sin \theta$. The chain rule givesTheir Euclidean norms are $\Vert {\partial }_r\Vert =1$ and $\Vert {\partial }_{\theta }\Vert =r$ for $r>0$.

Define $F(x)={\int }_a^{x}f(t)dt$. By the fundamental theorem of calculus, $F^{\prime }=f$. The chain rule givesIntegrating from $\alpha$ to $\beta$ and using $g(\alpha )=a$, $g(\beta )=b$ yields the change of variables formulaThe argument also covers a decreasing $g$ because both integrals are oriented.

In suffix notation,where ${\delta }_{ij}$ is the Kronecker delta. Since $r=(x_k{x}_k{)}^{1/2}$, the chain rule gives

Put $y(x)=w(z)$ with $z={\sin }^{2}x$. The chain rule givesand henceAfter substitution and division by $4z(1-z)$, the ordinary differential equation becomesComparison with the Gauss hypergeometric equation gives, up to interchanging $A$ and $B$,

At $z=0$ and $z=1$, the coefficient of $w^{\prime }$ has at most a simple pole and that of $w$ has at most a double pole, so both are regular singular points. The regular singular point at infinity criterion gives the same conclusion at infinity. The corresponding Papperitz symbol isIndeed, the exponent pairs are $(0,1-C)=(0,1/2)$ at zero, $(0,C-A-B)=(0,1/2)$ at one, and $(A,B)=(k/2,-k/2)$ at infinity; their sum also satisfies the Fuchs relation.

Underthe chain rule gives $dot{a}_j=(a_j/2){\dot{u}}_j$. Part (ii) and $a_j>0$ therefore implyConsequentlyThe conventions $u_0=u_n=-\infty$ give $e^{u_0}=e^{u_n}=0$, exactly matching $a_0=a_n=0$.

Along a solution, the chain rule and antisymmetry of $K$ givebecause $v^{T}Kv=0$ for every real vector $v$. ThereforeThe closed level curves around the strict minimum at the coexistence equilibrium are the periodic population cycles of the Lotka-Volterra predator-prey model.

With $\eta =x-t$ and $\xi =x+t$, the chain rule givesHenceso the wave equation is equivalent to $U_{\eta \xi }=0$. Integrating in each variable givesThe initial conditions implySolving and integrating yields the D'Alembert formula with initial velocity

If $|x-x_0|>r+t$, then both $x-t$ and $x+t$ lie outside the support interval $[x_0-r,x_0+r]$, and every point between them does as well. All three terms therefore vanish. Thuswhich is finite propagation speed.

For the finite string, differentiate its energy and use $y_{tt}=y_{xx}$:The fixed-end conditions hold for every $t$, so differentiation gives $y_t(0,t)=y_t(L,t)=0$. The boundary term vanishes and

The inverse Lorentz transformation of electromagnetic fields, from the primed rest frame to the unprimed frame, givesandEquivalently, $\mathbf{B}=\gamma \mathbf{v}\times {\mathbf{E}}^{\prime }/c^2$ for $\mathbf{v}=v{\mathbf{e}}_x$.

Using $1/{\mu }_0={ϵ}_0{c}^2$ and ${\gamma }^2=(1-v^2/c^2{)}^{-1}$,

The primed field is static, so $w$ depends on $t,x$ throughand has no explicit $t^{\prime }$ dependence. The chain rule givesConsequentlyThus the field-energy profile is transported rigidly with velocity $v{\mathbf{e}}_x$.

For fixed $(x_1,x_2)$, apply the given one-variable identity to the smooth functionThe chain rule givesThereforewhereEvery derivative of the integrand is continuous. Repeated differentiation under the integral sign therefore shows that $h$ is smooth. This is the Second-order Hadamard lemma.

Since $\beta =f^{-1}df$,The fundamental theorem of calculus and the chain rule then giveThus $F$ is constant.

Choose the positive local branch of the constraint,The chain rule givesand thereforeSubstitution into the original free energy gives exactly

For a planar equilibrium, multiply $\mu =0$ by ${\varphi }^{\prime }$ and use the chain rule. The result isMatching to either bulk phase fixes the constant:Consequentlyand the inverse profile is

The multivariable chain rule gives, at $(x,y)=(a+t\cos \gamma ,b+t\sin \gamma )$,Sufficient conditions for a strict local minimum at $t=0$ are $g^{\prime }(0)=0$ and $g^{\prime \prime }(0)>0$.

At a stationary point, $g^{\prime }(0)=0$ in every direction. Its second derivative is the quadratic form of the Hessian matrix. Ifthen completing the square givesfor every nonzero direction $(u,v)$. The Hessian is therefore positive definite, and the stationary point is a strict local minimum.

Since $r=(x_j{x}_j{)}^{1/2}$, the chain rule givesFor the radial vector field $g(r)\mathbf{x}$, the product rule and Einstein summation convention give

The implicit function theorem states that if $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a differentiable bijection with differentiable inverse $g$, the chain rule applied to $g\circ f$ and $f\circ g$ givesThus $D{f}_x$ is an isomorphism with inverse $D{g}_{f(x)}$.

If a continuously differentiable map $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has invertible derivative everywhere, the inverse function theorem makes it a local diffeomorphism. In particular it is an open map, so its image is open. Its image need not be closed: $x\mapsto \arctan x$ has nonzero derivative everywhere and image $(-\pi /2,\pi /2)$.

For the given map of elementary symmetric polynomials,and direct evaluation of the determinant givesHence the critical set isIts complement is the Zariski-open set on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A continuous path cannot change an ordering without crossing $C$. Therefore ${\mathbb{R}}^3\setminus C$ has exactlyconnected components.

For $N=1$, writeThe given evolution of the discrete scattering data givesSolving the resulting two scalar linear equations for the components of ${\psi }_1$ and substituting in the reconstruction formula givesChoosing the additive multiple of $2\pi$ so that $u\to 0$ as $X\to +\infty$ and integrating in $X$ yields the One-soliton solution of the sine-Gordon equation in light-cone coordinatesIt depends only on $X+T$ exactly when the two positive coefficients in the exponent agree:The unique positive solution is $l=1/2$, and then

The transformations satisfy $g^s{g}^r=g^{s+r}$, $g^0$ is the identity map, and $(g^s{)}^{-1}=g^{-s}$, so they form a one-parameter group. Ifthen the chain rule givesThus $g^s$ is a Lie point symmetry. Applied to the one-soliton family, it replaces $l$ by $l^{\prime }=e^{-s}l$. Taking $e^s=2l$ gives $l^{\prime }=1/2$, so every member is transformed to the function $F(X+T)$ found above.

For the stated $N=2$ solution, set $x=X+T$ and $t=T-X$. When $l^2+m^2=1/4$, its two arguments reduce toAt fixed $x$, only $\sin (2mt)$ varies, so the Sine-Gordon breather has fundamental period

Finally put $r^2=l^2+m^2$. Under the same symmetry, the parameters becomeand hence $(l^{\prime }{)}^2+(m^{\prime }{)}^2=e^{-2s}{r}^2$. Choosing $e^s=2r$ makes this sum $1/4$, proving that every solution in the family is symmetry-equivalent to the normalized breather.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Frechet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that