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Differentiate the energy and use the chain rule:After integration by parts, the middle term is $-\int (\Delta u)u_t$. Thereforeby the defocusing semilinear wave equation. Thus the total energy is conserved.

Write $X=\mu +\sigma Z$ with $Z$ standard normal and set $g(z)=f(\mu +\sigma z)$. The Gaussian Poincaré inequality and the chain rule giveHence $N(\mu ,{\sigma }^2)$ satisfies a $\sigma$-Poincaré inequality in the question's convention, where the constant is squared on the right-hand side.

Let $g:\mathbb{R}\to \mathbb{R}$ be continuously differentiable. Applying the Poincaré inequality for $p$ to $g\circ \varphi$ and using the chain rule givesThus the Pushforward of a Poincaré inequality by a Lipschitz function gives a $cL$-Poincaré inequality for $q$.

For these increasing warps, the square-root velocity function is $Q(h)(t)=\sqrt{h^{\prime }(t)}$. The chain rule givesThereforewhere the last equality again uses the change of variables formula. Taking square roots proves invariance under common right composition by $\gamma$.

Substitute $m_4=e^{\Psi }$ into the given equation. The chain rule givesThe incident reduced field is the constant one, so $m_4(-\Delta x)=1$ and $\Psi (-\Delta x)=0$. During a thin-screen crossing, $\Psi =O(\Delta x)$. Its mixed-derivative contribution integrates to $O(\Delta x)$. More precisely, its contribution to $\Psi$ is $O((\Delta x{)}^2)$, while the product of its first derivatives contributes only at still higher order. Thereforeand hence

The chain rule andgiveApplying the same rule to $L_1$ and $L_2$ in the Euler-Lagrange field equation yields the exact modulated equation

Multiply this equation by ${\Phi }_{\theta }$. Since $k$ and $\omega$ are independent of the fast phase,The product rule then rearranges the field equation into the exact modulated-wave first integralAverage this identity over one $2\pi$ period in $\theta$. Periodicity kills the first term, while differentiation of the averaged Lagrangian givesConsequently

The same two equations follow directly from the modulated variational principle. Varying $\Phi$ giveswhich is the exact field equation above. For a variation of $\Theta$, use $\delta k={\partial }_X\delta \Theta$ and $\delta \omega =-{\partial }_T\delta \Theta$. Integration by parts in $X$ and $T$ gives the averaged equation. Thus variation with respect to the periodic profile reproduces the local wave equation, whereas variation with respect to its slow phase gives the Whitham modulation equation for wave action.

Choose the orientation of a surface represented by a smooth Gauss map $N:S\to {\mathbb{S}}^2$. The shape operator at $p$ isIf its principal curvatures are $k_1,k_2$, then by definition

Let $E(x)=Ax+b$ be a proper Euclidean motion of Euclidean three-space, so $A\in SO(3)$. Orient $\tilde{S}=E(S)$ by transporting the normal:The chain rule gives the orthogonal conjugacyThe determinant and trace are invariant under similarity, so the Euclidean invariance of the shape operator yields

If the surface orientation is reversed, then $N$ and $S_p$ are multiplied by $-1$. Therefore the orientation reversal of surface curvature says

Hamilton's equations and the multivariable chain rule giveThus the Hamiltonian conservation law gives

For the phase-space dilation generator $D=q\cdot p$,Therefore the condition $\{f,D\}=0$ is the first-order equation

If $f(\lambda q,{\lambda }^{-1}p)=f(q,p)$ for every positive real $\lambda$, differentiate with respect to $\lambda$ at $\lambda =1$. The chain rule gives exactlyso every smooth scale-invariant $f$ satisfies the constraint. The same conclusion for negative $\lambda$ follows wherever the stated invariance is defined; differentiation only needs the positive component containing $\lambda =1$.

To say that $f$ is twice differentiable at $a$ means that $f^{\prime }$ is defined in a neighbourhood of $a$ and is differentiable at $a$; its derivative is $f^{\prime \prime }(a)$.

By the chain rule, near $a$,Since $f$ is differentiable and $g^{\prime }$ is differentiable at $f(a)$, another use of the chain rule shows that $g^{\prime }\circ f$ is differentiable at $a$. The product rule then proves that the displayed product is differentiable. Hence $g\circ f$ is twice differentiable, with the second derivative chain rule

Differentiability lets us writeand, with $k\to 0$,Substitute $k=f(a+h)-f(a)=f^{\prime }(a)h+r(h)$. Since $k=O(h)$, the remainder $s(k)$ is $o(h)$. Thusand the chain rule is

The space of test functions isA sequence ${\phi }_j$ converges to $\phi$ in $\mathcal{D}$ when all supports lie eventually in one compact set $K$ andfor every multi-index $\alpha$. A distribution is a linear functional $u:\mathcal{D}\to \mathbb{C}$ such that, for every compact $K$, there are $C_K>0$ and $m_K\in {\mathbb{N}}_0$ withwhenever $\mathrm{supp}\phi \subseteq K$. Convergence in ${\mathcal{D}}^{\prime }$ is pointwise convergence on test functions.

Continuity plainly implies sequential continuity. Conversely, suppose the linear form is sequentially continuous but the displayed estimate fails for some compact $K$. For every $j$, choose ${\phi }_j$ supported in $K$ such thatThen ${\phi }_j\to 0$ in $\mathcal{D}$, while $u({\phi }_j)\nrightarrow 0$, a contradiction. Hence the seminorm estimate holds on every compact set and $u\in {\mathcal{D}}^{\prime }$.

For a translation vector $h$ and a multi-index $\alpha$, defineThese definitions extend ordinary translation and differentiation to distributions.

If ${\tau }_{t{e}_i}u=u$ for every $t$, differentiating its pairing at $t=0$ gives ${\partial }_{i}u=0$. Conversely, if ${\partial }_{i}u=0$, then for every test functionThe pairing is constant in $t$, hence ${\tau }_{t{e}_i}u=u$. This proves both directions of translation invariance and vanishing distributional derivative.

Finally, the distributional differentiation obeys the linear chain rule under the linear coordinates $s=x-y$ and $t=x+y$. Thusand likewiseas distributions. Adding the two identities givesthe one-dimensional wave equation; this is the low-regularity form of the travelling waves in the D'Alembert formula.

Use the reflection extension from a half-spaceIt is plainly linear and restricts to $u$ on $U$. For a smooth $u$, the chain rule givesA change of variables therefore givesfor $p<\infty$, with the evident equality of essential suprema for $p=\infty$. Approximate a general $u$ by the smooth functions from part c. The estimate makes their reflections Cauchy in $W^{1,p}({\mathbb{R}}^2)$, and their limit defines a bounded Sobolev extension operator with $\Vert E\Vert \le 2^{1/p}$.

For $s\ge 0$ and $\xi \in \mathbb{R}$, define the characteristic curve $X(t;s,\xi )$ byThe bounded derivative $F_x$ makes $F(t,\cdot )$ globally Lipschitz, uniformly in $t$. On each finite time interval, $|F(t,x)|\le |F(t,0)|+L|x|$, so Gronwall inequality prevents finite-time escape. The Picard-Lindelof theorem therefore gives a unique trajectory for every finite $t\ge 0$. Differentiation in $\xi$ givesso the characteristic flow map is a $C^1$ increasing diffeomorphism.

Along a characteristic, the chain rule changes the equation intoTracing $(t,x)$ backward by the flow therefore givesThe regularity of the flow makes this a $C^1$ classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.

At $x\in \Sigma$, keep the unit normal $N(x)$ fixed and define the $j$th normal derivative byFor $j=1$, the chain rule and $N=\nabla \varphi /|\nabla \varphi |$ give

The characteristic flow map solves the ordinary differential equationand henceAlong this characteristic curve, the chain rule givesThe value is therefore constant, and tracing $(t,x)$ back to time zero gives the classical solutionDirect differentiation verifies both the linear transport equation and its initial value.

For $\gamma \ne 0$ and $x>0$, the change of variables $u=x^{\gamma }$ convertsinto $Y_{uu}+Y=0$, where $y(x)=Y(u)$. The chain rule cancels the first-derivative terms. Thus $\cos (x^{\gamma })$ and $\sin (x^{\gamma })$ form a pair of linearly independent solutions, with Wronskian $\gamma x^{\gamma -1}$. This is a variable-speed reparametrization of the harmonic oscillator equation.

If $\mu$ has Poincaré constant $C$ and $\varphi$ is $L$-Lipschitz, then the pushforward measure ${\varphi }_{\ast }\mu$ has Poincaré constant at most $C{L}^2$. Apply the original inequality to $g\circ \varphi$ and use the chain rule.

For $t>0$, the change of variables $\tau =1/t$ and $u(t)=v(\tau )/\tau$ converts $t^4{u}^{\prime \prime }+{\lambda }^{2}u=0$ into $v^{\prime \prime }+{\lambda }^{2}v=0$. The chain rule gives $u^{\prime }=v-\tau v^{\prime }$ and $u^{\prime \prime }={\tau }^3{v}^{\prime \prime }$. Thus $u=t[A\cos (\lambda /t)+B\sin (\lambda /t)]$. The prefactor cancels the derivative term introduced by inverting the independent variable.

For hidden activations $h_j=\sigma (b_j+{\sum }_r{W}_{jr}{x}_r)$ and class probabilities $p_c={\mathrm{softmax}}_c(a+Vh)$, a one-hot label $t$ has log-likelihood $\ell ={\sum }_c{t}_c\log p_c$. Put ${\delta }_c=t_c-p_c$ and ${\Delta }_j=h_j(1-h_j){\sum }_c{V}_{cj}{\delta }_c$. ThenThese follow from the chain rule and give single-observation stochastic gradient descent updates for the negative log-likelihood.