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Expand about $t=t_{n+2}$ and use $y^{\prime \prime }=f^{\prime }(y)f(y)$ from the chain rule. The residual of the exact solution isThe coefficients through $h^3$ vanish, and the displayed fourth-order coefficient does not. Thus the multiderivative multistep method has order three.

At $h=0$, its first characteristic polynomial isIts roots are $1$ and $1/7$, with the unit-modulus root simple, so it satisfies the root condition for a multistep method and is zero-stable. Combining this with the defect estimate proves convergence of order three, assuming sufficiently smooth $f$, order-three starting values, and the nearby branch of each implicit update.

More explicitly, the right-hand side is $h{F}_h(y_{n+2})$, where $F_h=6f/7-2h{f}^{\prime }f/7$ has a uniform local Lipschitz continuity bound for small $h$. A bounded zero-stable impulse response and the discrete Gronwall inequality therefore control the accumulated $O(h^4)$ defects by $O(h^3)$ over fixed time intervals. This is convergence of a zero-stable multiderivative method, rather than a direct application of the first-derivative-only Dahlquist equivalence theorem.

Restrict $\phi$ to an arbitrary line segment by defining $q(t)=\phi ((1-t)x+ty)$ for $0\le t\le 1$. The chain rule and the assumed positive semidefinite matrix condition on the Hessian matrix giveThus $q^{\prime }$ is non-decreasing. For $0<t<1$, integrating $q^{\prime }$ on the two subintervals givesRearranging, and including the immediate endpoint cases, yieldsThis is precisely the definition of a convex function.

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.