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The Leibniz integral rule combines contributions from both moving endpoints with differentiation under the integral sign:The partial derivative in the last integral holds $t$ fixed. Suitable regularity and convergence are assumed as stipulated.

For the Leibniz integral rule, the upper endpoint has derivative $2\pi x$ and the lower endpoint has derivative $1$. The integrand has partial derivative $4xt+\sin t$. ThereforeThe last integral is $[2x{t}^2-\cos t{]}_x^{\pi x^2}=2{\pi }^2{x}^5-2x^3-\cos (\pi x^2)+\cos x$. Combining terms gives

The gradient is horizontal when $h_y=0$ and vertical when $h_x=0$. Factoring those partial derivatives gives the two loci:Thus each locus consists of an axis and two parabolas. Their common points are the five stationary points found below. At those points the gradient is zero and has no direction, so they are included in both algebraic zero-component conditions but excluded if “parallel” is required to mean a nonzero directed vector.

Put $q=x^{2}yz$. The chain rule and product rule giveSumming these partial derivatives gives the Laplacian

For the given differential form, set $P=x(y^4+1)$ and $Q=y(2x^2{y}^2+1)$. Their partial derivatives satisfyIn fact an explicit potential issince $F_x=P$ and $F_y=Q$. Thus the differential form is an exact differential form on all of ${\mathbb{R}}^2$.

Replacing $y^2$ by $u$ suggests the potentialDirect differentiation verifies that $dH/dx$ is exactly the left-hand side of the required ordinary differential equation. This direct verification remains valid even for negative $u$; it does not require a real square root $y=\sqrt{u}$. ThereforeThe initial condition gives $C=1$. Solving this quadratic equation in $u$ and selecting the branch that has $u(1)=0$ givesThe second expression also defines the removable value $u(0)=1$. The maximal open interval containing $1$ on which this is a differentiable real solution isAt either endpoint the square root vanishes and $2x^{2}u+1=0$, while $x(u^2+1)\ne 0$, so the original ordinary differential equation cannot hold there with a finite derivative.

Translate the coordinates by $u=x-1$ and $v=y+1$, and write $E=e^{-(u^2+v^2)/2}>0$. The partial derivatives areIf $u=0$ or $v=0$, these equations together force $u=v=0$. Otherwise they require $u^2=v^2=1$. Thus the complete set of stationary points is

Apply ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ once more. The chain rule and symmetry of second derivatives giveHere and in the next part the second partial derivatives are understood with the usual smoothness required for this partial differential equation.

The coordinate partial derivatives are ${\xi }_y=1/\sqrt{y}$ and ${\eta }_y=-1/\sqrt{y}$. The multivariable chain rule therefore givesSince $\xi -\eta =4\sqrt{y}>0$, the expression entirely in the new coordinates isThe inverse coordinate formulas are $x=(\xi +\eta )/2$ and $y=(\xi -\eta {)}^2/16$, with domain $\xi >\eta$.

Substitute the partial derivatives obtained above. The first-order terms cancel exactly:The transformed partial differential equation is thereforeIn these characteristic coordinates, the mixed partial derivative is the only surviving term.

Put $\mathbf{W}=\nabla \times \mathbf{F}$ and $\mathbf{v}=\dot{\mathbf{r}}$. All vector fields and their spatial partial derivatives below are evaluated at $\mathbf{r}(t)$. The product rule for a dot product givesBecause the vector field has no explicit time dependence, the multivariable chain rule gives $d\mathbf{W}/dt=J_{\mathbf{W}}\dot{\mathbf{r}}$, where the Jacobian matrix has entries $(J_{\mathbf{W}}{)}_{ij}={\partial }_j{W}_i$. ConsequentlyThe first term measures the change of the curl sampled as the point moves; the second measures the change in the velocity vector used in the dot product.

To give the result explicitly in $P,Q,R$, define the directional derivative operator $L=\dot{x}{\partial }_x+\dot{y}{\partial }_y+\dot{z}{\partial }_z$. Sincethe same expression isFor example $L(R_y-Q_z)=\dot{x}(R_{yx}-Q_{zx})+\dot{y}(R_{yy}-Q_{zy})+\dot{z}(R_{yz}-Q_{zz})$; the other two terms expand in the same way. Thus only the first and second spatial partial derivatives of $P,Q,R$, together with the velocity and acceleration of the parametric curve, occur.

Write $\Psi (\eta ,\xi )=\psi (x,t)$. The inverse coordinate formulas $x=(\eta +\xi )/2$ and $t=(\eta -\xi )/(2c)$ are valid because $c>0$. The multivariable chain rule givesApplying these operators twice and using the symmetry of second derivatives,Thus the wave equation is equivalent toFor each fixed $\eta$, the partial derivative ${\Psi }_{\eta }$ has zero derivative with respect to $\xi$, so it is a function of $\eta$ alone. Taking an antiderivative in $\eta$ gives $\Psi =f(\eta )+g(\xi )$, where the integration constant may depend on $\xi$. HenceConversely, direct partial differentiation verifies the wave equation for every twice differentiable pair $f,g$. The two terms describe profiles moving left and right respectively at wave speed $c$ along the characteristic lines for the one-dimensional wave equation.

The divergence sums the corresponding partial derivatives:

The radial component of the curl in spherical coordinates isIt contains no $f_r$, so $R$ cannot contribute. Substituting the tangential components givesThe mixed partial derivatives of smooth $G$ agree, so its terms cancel. Multiplication by $r$, to take the dot product with the position vector $\mathbf{r}=r{\mathbf{e}}_r$, does not change this independence from $R,G$.

Differentiate the Cauchy-Riemann equations and use equality of mixed partial derivatives, which holds because holomorphic functions are smooth:Thus both parts are harmonic functions.

The Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. A holomorphic function is smooth, so equality of mixed partial derivatives givesThus its real and imaginary parts are harmonic functions.

An entire function is holomorphic at every finite point of the complex plane. The globally convergent power series for the exponential implies that $e^z$ and $e^{-z}$ are entire. Their difference therefore givesat every complex $z$, proving that the hyperbolic sine is entire. For powers, the base case $f_1(z)=z$ has derivative one everywhere. If $z^n$ is entire with derivative $n{z}^{n-1}$, the product rule gives $(z^{n+1}{)}^{\prime }=z^n+zn{z}^{n-1}=(n+1)z^n$ everywhere. Induction proves that every positive integer power is entire.

A function has a pole of order $n$ at $z_0$ when it is holomorphic in a punctured neighbourhood and $g(z)=(z-z_0{)}^{n}f(z)$ extends holomorphically to $z_0$ with $g(z_0)\ne 0$. An isolated essential singularity is neither a removable singularity nor a pole; equivalently its Laurent series has infinitely many nonzero negative-power terms. In particularhas an essential singularity at zero.

For $f_N$, the Laurent series about infinity isTo classify the point correctly, use $w=1/z$. Then $f_N(1/w)=w^{-N}{e}^w$, and $e^w$ is holomorphic with value one at zero. ThereforeThe pole at infinity is compatible with an essential singularity at the finite point zero. Negative powers of $z$ are regular powers of the local coordinate $w$ near infinity, so they do not make that point essential.

Since $\mathbf{V}=\nabla u$, its divergence is the Laplacian of the velocity potential. Using the Cauchy-Riemann equations and equality of mixed partial derivatives givesThus this is an incompressible flow, and $u$ is a harmonic function.

Differentiate the Cauchy-Riemann equations and use equality of mixed partial derivatives:Thus the real part is a harmonic function. The same calculation gives $\Delta v=0$. Every holomorphic function has the required smoothness.

Work in a Cartesian coordinate system, write ${\partial }_j=\partial /\partial x_j$, and use the Einstein summation convention. For a twice continuously differentiable vector field, the contraction of two Levi-Civita symbols givesHere the Kronecker deltas select components, and equality of mixed partial derivatives gives the curl of the curl identityFor the second identity, the same contraction followed by the product rule yieldsThus the curl of a cross product isThe directional derivatives in the last two terms act on the vector field immediately to their right.

For real-direction increments, complex differentiability gives $f^{\prime }=u_x+i{v}_x$. For imaginary-direction increments it gives $f^{\prime }=(u_y+i{v}_y)/i=v_y-i{u}_y$. Equating these expressions gives the Cauchy-Riemann equationsA holomorphic function is smooth, so mixed partial derivatives commute. Differentiating these equations givesThus both real and imaginary parts are harmonic functions:

Complex differentiability requires the same limit for increments along the real and imaginary directions. Real increments give $f^{\prime }=u_x+i{v}_x$. Imaginary increments giveEquating real and imaginary parts yields the Cauchy-Riemann equationsThey hold throughout the open region where $f$ is holomorphic. Conversely, continuous first real partial derivatives satisfying these equations on an open set ensure holomorphy there.