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For constant $m$, the curl of a cross product identity reduces toCommuting mixed second partial derivatives gives $(m\cdot \nabla )\nabla \varphi =\nabla (m\cdot \nabla \varphi )$. Hence the requested sum equals $m\Delta \varphi$, which is zero because $\varphi$ is harmonic. The argument assumes the indicated second derivatives are continuous on the domain.

From the preceding gradient, ${\partial }_x^2(r^{-1})=-r^{-3}+3x^2{r}^{-5}$. Summing the three Cartesian second partial derivatives givesThus the reciprocal radius is harmonic away from the origin. Its negative gradient isThe curl vanishes because $E$ is a gradient field and mixed second derivatives commute. The exclusion of the origin is essential: the potential and field are singular there.

If $Pdx+Qdy=dF$ with continuous second partial derivatives of $F$, then $P=F_x$, $Q=F_y$. The symmetry of second derivatives requiresThis is a necessary condition for an exact differential. With continuously differentiable coefficients it is also locally sufficient, and is globally sufficient on a simply connected open domain; a domain with holes can require additional global information.

The partial derivatives are $f_x=e^{xy}(1+xy)$ and $f_y=x^2{e}^{xy}$. At $(1,1)$ they have values $2e$ and $e$. The first-order Multivariate Taylor polynomial therefore gives the changeThus the requested linear approximation is $e(2\delta x+\delta y)$; it is the change from $f(1,1)=e$, not the new function value itself.

The first partial derivatives factor asBoth vanish at the origin, verifying that it is a stationary point. Since $2x^2+1>0$, every stationary point has $y=0$; the first equation then gives $x=0,\pm 1$. Thus one requested additional point is $\boxed{(1,0)}$; the other is $\boxed{(-1,0)}$.

If the form is $dF$ for a scalar function with continuous second partial derivatives, then $p=F_x$ and $q=F_y$. Equality of the mixed partial derivatives therefore requiresThis is necessary for an exact differential. For continuously differentiable coefficients on a simply connected open domain it is also sufficient; on a domain with holes, the equality alone need not give a single-valued global potential.

In Cartesian coordinates, the divergence uses the matching coordinate for each component:Each of these three partial derivatives vanishes.

The curl of a gradient vanishes because the mixed second partial derivatives of the smooth function $\varphi$ agree. Therefore $\nabla \times \mathbf{v}=\nabla \times \nabla \varphi =0$ everywhere, andUnlike part (i), this integrand does not involve the problematic auxiliary field.

Let $M=x+y$ and $N=x$. Their cross partial derivatives agree: $M_y=N_x=1$. More directly, construct the scalar functionThis exhibits the expression as an exact differential. Along a solution of the differential equation,Thus the general solution isFor $C=0$, the graph $y=-x/2$ also extends smoothly through $x=0$ and satisfies the original, undivided equation there. Nonzero $C$ cannot extend through zero as a finite-valued differentiable solution. The extra vertical component $x=0$ of the implicit zero-level equation is not a graph $y(x)$ on an interval and is not an additional solution of this ordinary differential equation.

The product rule gives the two partial derivativesThey both vanish at the origin. The exponential function is never zero, so any further stationary point must satisfy $x=0$ or $x^2-y^2=2$, and $y=0$ or $y^2-x^2=2$. Having both coordinates nonzero would require both opposite differences to be $2$, which is impossible. If $y=0$, the nonzero choices are $x=\pm \sqrt{2}$; if $x=0$, they are $y=\pm \sqrt{2}$. Hence the full list is

Apply the fundamental theorem of calculus separately to the upper and lower limits. No evaluation of the integral is needed:The first result has no $y$ dependence and the second no $x$ dependence, so the mixed partial derivative is

Write the differential one-form as $Mdx+Ndy$. An exact differential form must satisfy equality of mixed partial derivatives, $M_y=N_x$. HereEquality at every $(x,y)$ holds precisely when $\boxed{a=2}$. Since these polynomials are smooth on the simply connected plane, the closed-form condition is also sufficient for exactness; the explicit potential of a conservative vector field constructed in (c) verifies that sufficiency directly.

With $a=2$, integrate the $dx$ coefficient with respect to $x$:Its other partial derivative is ${\Phi }_y=x^2/2-2xy+g^{\prime }(y)$. Matching the $dy$ coefficient gives $g^{\prime }=0$. Therefore the differential equation is $d\Phi =0$, and its implicit general solution isThese are level curves of the potential of a conservative vector field. Where $N\ne 0$, they give the usual graph solution $dy/dx=-M/N$. At $N=0$, the implicit description retains possible vertical tangents instead of dividing by zero.

Write the kernel as $K(s,t)=(4\pi t{)}^{-1/2}{e}^{-s^2/(4t)}$, with $t>0$. Its partial derivatives areThus the heat kernel itself satisfies $K_t=K_{ss}$. The proposed convolution consequently givesThis verifies the heat equation. The parameter is positive time; the real integral formula is not defined at negative time.

The passage of derivatives through the integral needs justification. A sufficient condition is the Gaussian-weighted absolute integrability in heat-kernel convergence for Gaussian-integrable initial data; it holds, in particular, for data of polynomial growth. On compact positive-time intervals strictly below an integrable later time, the differentiated kernels have an integrable Gaussian bound, so differentiation under the integral sign applies.

Even if the PDF's assumed convergence is read as ordinary conditional convergence, one can justify the same calculation. Choose a later time $T>t$ and let $H(\xi )$ be a primitive of $\phi (\xi )e^{-{\xi }^2/(4T)}$. Convergence at spatial point zero and time $T$ makes $H$ bounded on both tails. Put $q(x,\xi ,t)=K(x-\xi ,t)e^{{\xi }^2/(4T)}$. This function, its spatial and time derivatives, and their $\xi$-derivatives decay as Gaussians when $t<T$. Integration by parts rewrites the convolution as $-\int H{q}_{\xi }d\xi$, with vanishing endpoint terms. Boundedness of $H$ now permits differentiation under this absolutely convergent integral; since $q_t=q_{xx}$, the same equation follows. Thus one need not silently equate conditional and absolute convergence.

The partial derivatives are $f_{xx}=4f$ and $f_t=\omega f$. Since the exponential function never vanishes, substituting into the partial differential equation gives $4=6\omega$. Hence

Assume the components of $\mathbf{F}$ have continuous second partial derivatives. Expanding the divergence of its curl givesEach positive term cancels a negative term because the corresponding mixed derivatives commute. Consequently

First take $F_y=x{e}^{\sin (xy)}\cos (xy)$, then differentiate with respect to $x$. The product rule and chain rule giveThe function is smooth, so the two mixed partial derivatives agree.

The partial derivatives give $x^3=9y$ and $y^3=9x$. If either coordinate is zero, both vanish. Otherwise substitution gives $x^8=3^8$, so the real stationary points are $(0,0)$, $(3,3)$ and $(-3,-3)$.

The Hessian matrix isAt the origin its eigenvalues are $36$ and $-36$, so it is a saddle at level zero. At either of the other points its eigenvalues are $72$ and $144$, so both are strict minima at level $-162$. In fact,with equality only at those two points. HenceFor levels $-162<c<0$, two separate closed contours surround the two minima. At $c=0$ the contours form two lobes, in the first and third quadrants, meeting at the origin. Their limiting tangent directions there are the coordinate axes because the leading term is $-36xy$. On the diagonal the outer zero-level points are $(\pm 3\sqrt{2},\pm 3\sqrt{2})$. Positive levels form one closed outer contour enclosing both wells; there are no contours below $-162$. The graph is symmetric under exchanging $x,y$ and under reversing both signs, but not under a quarter-turn. These features are the $a=3$ case of the quartic double well with bilinear coupling.

Take the Cartesian partial derivatives component by component. The divergence isThe constant third component has zero derivatives. The curl consequently has only two nonzero components:

Stationarity requires both first partial derivatives to vanish:For example, choosing $x=\pi$ and $\cos y=0$ gives two points strictly inside the prescribed rectangle:Both satisfy the two derivative equations directly.