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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

Put $S(\mathbf{x})=\mathbf{x}\cdot \mathbf{A}\cdot \mathbf{x}$. Since $\mathbf{u}=-S{r}^{-5}\mathbf{x}$ and $\nabla \times \mathbf{x}=0$, the product rule givesFor the symmetric stresslet tensor $\mathbf{A}$,The final term is parallel to $\mathbf{x}$ and drops out of the cross product, leaving

Vary $g^{ac}{g}_{cb}={\delta }_b^a$. The product rule givesand multiplication by $g^{bd}$ yields the metric variation

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: