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To first order in the weak fluctuation, $n^2-1=2\mu W+O({\mu }^2)$, soForintroduce the signs $s=(1,-1,1,-1)$. The product rule gives

Assume that $W$ is a zero-mean stationary Gaussian random field, that its longitudinal correlation length is short compared with the envelope's evolution scale, and that the propagation distance is long compared with that correlation length. The forward Markov approximation then neglects diffraction during one correlation length. Applying the Furutsu–Novikov formula closes the last average at second order in $\mu$. Define the integrated longitudinal autocorrelation function of a random fieldThenwhereWriting $C_{ij}=C(z_i-z_j)$ and using the evenness of the covariance gives the equivalent expressionThe derivation also assumes paraxial propagation, weak scattering, sufficient regularity to interchange differentiation and expectation, and statistical homogeneity in both coordinates. Without the short-correlation approximation, the Gaussian identity produces a nonlocal longitudinal memory integral rather than this local closed equation.

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.

WriteDivide by $h$ and let $h\to 0$. Differentiability implies continuity, so $g(a+h)\to g(a)$. Therefore the product rule gives

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

Use the Polynomial representation of the Heisenberg Lie algebra on the infinite-dimensional polynomial ring $\mathbb{C}[x]$:The product rule gives $[d/dx,x]=1$, so this is a Lie algebra representation. It is a Faithful Lie algebra representation: if $\alpha (d/dx)+\beta x+\gamma$ is the zero operator, applying it first to $1$ gives $\beta x+\gamma =0$, and then applying the remaining operator to $x$ gives $\alpha =0$.

To prove irreducibility, let $W$ be a nonzero invariant subspace and choose a nonzero polynomial in $W$ of least degree. If its degree were positive, repeated differentiation would produce a nonzero element of smaller degree, so $W$ contains a nonzero constant. Invariance under multiplication by $x$ then puts every monomial $x^n$ in $W$, and hence $W=\mathbb{C}[x]$.

Because $v$ is a divergence-free vector field, the product rule givesThe integral of a divergence over the periodic domain vanishes. Hence integration by parts yieldsThis is the skew-symmetry of incompressible transport.

The regularity from part (i) makes $u\omega$ and its first weak derivatives square-integrable. The product rule therefore holds in the distributional derivative sense:Since $u={\nabla }^{\perp }\Psi$, equality of mixed weak derivatives gives $\nabla \cdot u=0$. Both remaining expressions belong to $L_{\mathrm{per}}^2$, so their distributional equality is an equality in that space:

On $(0,1)$, the product rule and chain rule give $f^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$. On $(-1,0)$, the function is constant and has derivative zero. At zero,Its absolute value is at most $|h|$, so the squeeze theorem gives $f^{\prime }(0)=0$. ThereforeFor $x_n=1/(2\pi n)$, we have $x_n\to 0$ but $f^{\prime }(x_n)=-1$ for every $n$. This differs from $f^{\prime }(0)=0$, so $f^{\prime }$ is not continuous at zero. Indeed, $y_n=1/((2n+1)\pi )$ gives $f^{\prime }(y_n)=1$, so the right-hand limit of a function does not exist. Thus $f$ is differentiable everywhere on $(-1,1)$, while $f^{\prime }$ is discontinuous at zero.

Choose $\mathbf{F}=\nabla \psi$ in the preceding product rule and apply the divergence theorem. Since $\nabla \cdot \nabla \psi =\Delta \psi$, this gives the first of Green's identities:Here ${\partial }_n\psi =\mathbf{n}\cdot \nabla \psi$ is the outward normal derivative. Interchange $\varphi$ and $\psi$, then subtract. The gradient dot products cancel, yielding the second of Green's identities:For example, $\varphi ,\psi \in C^2(\overline{V})$ with the needed extensions suffice for these applications of the divergence theorem.

Apply the product rule to each vector component and sum:Thus the divergence of a scalar field times a vector field contains a divergence term and a dot product with the gradient of the scalar field.

In Cartesian coordinates, write the constant vector as $\mathbf{c}=(c_1,c_2,c_3)$. The product rule and constancy of its components giveThe last expression is the dot product with the gradient, so

No. On $(-1,1)$ take $c=0$ andAt zero, its difference quotient has absolute value $|x\sin (1/x)|\le |x|\to 0$, so $f^{\prime }(0)=0$. Away from zero, the product rule and chain rule apply. DefineThese are distinct positive inputs tending to zero. Since $\sin A_n=0$ and $\sin (A_n+B_n)=\sin B_n$,The even-indexed subsequence tends to $-2/\pi$, whereas the odd-indexed subsequence tends to $2/(3\pi )$. Thus the limit of a sequence does not exist. The two points lie on the same side of $c$, so the nonnegative-weight estimate in part (b) is unavailable. This is a concrete instance of differentiability does not imply continuity of the derivative.

In Cartesian coordinates, a parametrization of the curve isDifferentiate using the product rule:Squaring and adding cancels the mixed terms, giving $(x^{\prime }{)}^2+(y^{\prime }{)}^2=(f^{\prime }{)}^2+f^2$. The arc length is the integral of the speed, soThis assumes the parametrization is piecewise continuously differentiable and counts length with multiplicity if the curve is retraced.

For any fixed subregion $\Omega$, with no internal heat generation, conservation of energy says that the rate of stored thermal energy equals the incoming heat flux:The minus sign is needed because $\mathbf{q}\cdot \mathbf{n}>0$ denotes outgoing heat flux. Substitute Fourier's law $\mathbf{q}=-k\nabla T$ and apply the divergence theorem:Here differentiation under the integral uses a fixed region and sufficient regularity of $T$. Since the identity holds for every $\Omega$, continuity of the integrands gives the local heat equationFor spatially varying thermal conductivity, the product rule gives $\nabla \cdot (k\nabla T)=k{\nabla }^{2}T+\nabla k\cdot \nabla T$; the second term cannot generally be discarded.

Suppose $T^{(1)}$ and $T^{(2)}$ are two sufficiently regular steady state solutions with the same prescribed temperature on the boundary. Their difference $w=T^{(1)}-T^{(2)}$ satisfiesThe identical PDE source terms cancel. Apply the product rule and the divergence theorem to $wk\nabla w$; this is the weighted form of Green's first identity:Since $k>0$, the nonnegative integrand must vanish, so $\nabla w=0$. Hence $w$ is constant on each connected component of $R$, and its zero Dirichlet boundary condition makes each constant zero. Therefore $\boxed{T^{(1)}=T^{(2)}}$: the solution, if it exists, is unique. This proves Dirichlet uniqueness for variable-conductivity heat conduction; it does not require a constant thermal conductivity.

For $r>0$, the chain rule gives $\partial r/\partial x_i=x_i/r$. Applying the product rule to the components of the vector field yieldsThe calculation is pointwise away from the origin; any extension of the vector field to the origin must be considered separately.

Write ${\partial }_i=\partial /\partial x_i$ and use $\times$ for the paper's wedge notation, which denotes the cross product. The given expression for the curl and the product rule giveApply the vector triple product identity $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})=\mathbf{b}(\mathbf{a}\cdot \mathbf{c})-\mathbf{c}(\mathbf{a}\cdot \mathbf{b})$ separately to the two terms:Summing over the Cartesian basis vectors identifies the two directional derivatives and the two divergences:Here, for example, $((\mathbf{w}\cdot \nabla )\mathbf{v}{)}_j={\sum }_i{w}_i{\partial }_i{v}_j$. This proves the curl of a cross product directly from the formula supplied.

Apply the product rule inside the Cartesian expression for the curl:The first sum is the gradient of the scalar field and the second is the curl of the vector field. Thus the curl product rule for a scalar field is

All time derivatives in this part are taken relative to the rotating reference frame, so its basis is held fixed during differentiation. Put $f(\varphi )=p(\varphi )\sin \varphi$ and $q(\varphi )=p(\varphi )\cos \varphi$. The chain rule and product rule giveExplicitly,The separate Coriolis force and centrifugal force terms in the stated equation account for the motion of the basis relative to an inertial observer.

For polar coordinates, the rotating unit vectors satisfy ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$. Applying the product rule twice to $\mathbf{r}=r{\mathbf{e}}_r$ gives the velocity and accelerationNewton's second law for a central force now givesMultiplying the azimuthal force equation by $r/m$ yieldsThis is conservation of specific angular momentum. Here $F(r)$ is the signed radial component of the force; a negative value points inward.

When $h\ne 0$, $\theta$ is a valid local parameter along the trajectory. Put $u=1/r$ and let primes denote derivatives with respect to $\theta$. The chain rule yieldsSubstituting into the radial Newton's second law equation and dividing by $-m{h}^2{u}^2$ proves the Binet equation: