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Differentiate the position vector with respect to the three coordinates to obtain the coordinate basis vectors:The chain rule gives $d\mathbf{x}={\mathbf{h}}_{u}du+{\mathbf{h}}_{v}dv+{\mathbf{h}}_{\varphi }d\varphi$. The scale factors of orthogonal coordinates are the corresponding norms:Direct dot products give ${\mathbf{h}}_u\cdot {\mathbf{h}}_v=uv-uv=0$, ${\mathbf{h}}_u\cdot {\mathbf{h}}_{\varphi }=0$, and ${\mathbf{h}}_v\cdot {\mathbf{h}}_{\varphi }=0$. Thus the parabolic coordinates are orthogonal wherever the coordinate vectors are nondegenerate. Their degeneracy on $uv=0$ is the usual axis degeneracy of the azimuthal coordinate.

The Fermat principle makes the optical path length stationary. For a graph $y=\xi (x)$ the variational integrand isThe Euler-Lagrange equation isUse the chain rule $dn(x,\xi (x))/dx=n_x+n_y{\xi }^{\prime }$ and differentiate ${\xi }^{\prime }/\sqrt{1+{\xi }^{\prime 2}}$ to obtainMultiplying through and combining the $n_y$ terms gives the required second-order ordinary differential equation with no derivatives in denominators:All partial derivatives of the refractive index are evaluated along the ray.

Put $Y(\theta )=y(\cos \theta )$, with $0<\theta <\pi$. The chain rule givesConsequently the original differential expression becomes $-Y^{\prime \prime }-2\cot \theta Y^{\prime }$. If $Y=F/\sin \theta$, direct differentiation givesTaking $F=\sin n\theta$ therefore yieldsThe apparent endpoint singularities are removable. Expansion at $\theta =0,\pi$ givesThus both endpoint regularity conditions hold. This is the Chebyshev polynomial of the second kind $U_{n-1}(x)$.

The three coordinate basis vectors are the partial derivatives of the position vector with respect to the three coordinates. Direct differentiation givesThe multivariable chain rule then gives $d\mathbf{x}={\mathbf{h}}_{\xi }d\xi +{\mathbf{h}}_{\theta }d\theta +{\mathbf{h}}_{\varphi }d\varphi$. These are coordinate tangent vectors, not necessarily unit vectors.

Use a logarithmic change of variable in a Sturm-Liouville problem: set $t=\log x$ and $Y(t)=y(e^t)$. The chain rule givesThe endpoint conditions become $Y(0)=Y(1)=0$. Multiplying $-Y^{\prime \prime }=(\lambda +1)Y$ by $Y^{\ast }$ and integrating shows that $\lambda +1>0$ for a nonzero solution. Put $\mu =\sqrt{\lambda +1}$. The first endpoint selects $Y=A\sin (\mu t)$ and the second requires $\mu =n\pi$, $n\ge 1$. HenceThese eigenfunctions are orthonormal in the specified weighted inner product, because $w(x)dx=dx/x=dt$ and

Choose a standard mollifier $\rho \in C_0^{\infty }({\mathbb{R}}^n)$ with $\rho \ge 0$, $\int \rho =1$ and support in the unit ball, and set ${\rho }_{\epsilon }(z)={\epsilon }^{-n}\rho (z/\epsilon )$. Define the convolutionThe smoothing properties of a mollifier give $u_{\epsilon }\in C^{\infty }$ and ${\partial }_j{u}_{\epsilon }={\rho }_{\epsilon }\ast D_{j}u$. Apply the chain rule to $t\mapsto u_{\epsilon }(y+t(x-y))$ and integrate its derivative by the fundamental theorem of calculus:

The identity theorem states that if two holomorphic functions on a connected open set agree at a sequence of distinct points with an accumulation point inside that set, then they agree everywhere on it. Equivalently, the zeros of a holomorphic function which is not identically zero cannot accumulate inside its domain.

Because $f,g$ never vanish, their logarithmic derivatives are holomorphic. The holomorphic functionvanishes at every $a_n$. Since the distinct $a_n$ converge to $a\in U$, the identity theorem gives $H\equiv 0$. Applying the quotient rule now givesA holomorphic function with zero derivative on a domain is constant. One can see this locally from its Taylor series, then propagate the constant using connectedness. Hence $f/g=c$ on $U$, with $c\ne 0$, andThe conclusion can fail if the accumulation occurs outside $U$. Take the punctured complex plane $U=\mathbb{C}\setminus \{0\}$ andBoth functions are nowhere vanishing and holomorphic on $U$, and $a_n\to 0\notin U$. The chain rule givesso these logarithmic derivatives agree at every $a_n$. However, $f$ is not constant: for example $f(1/\pi )=e^{-1}$ and $f(1/(2\pi ))=e$. Therefore no constant $c$ gives $f=cg$.

On the slit domain $D$, choose the complex argument $\theta ={\mathrm{Arg}}_{(0,2\pi )}z$. The resulting holomorphic logarithm is $L(z)=\log |z|+i\theta$, with $L^{\prime }(z)=1/z$. DefineIt is holomorphic as a composition of holomorphic functions, and $f(z{)}^2=e^{L(z)}=z$. At $z=-1$ the chosen argument is $\pi$, so $f(-1)=i$.

Suppose a holomorphic square root $g$ existed on the whole punctured complex plane. It would be nonzero, and differentiating $g^2=z$ would give $2g{g}^{\prime }=1$. There are several ways to express the resulting obstruction; a direct one uses the unit circle. SetThe chain rule and $g^{\prime }(z)=g(z)/(2z)$ imply $q^{\prime }(t)=0$, so $q$ is constant. ButThese cannot agree because $g(1{)}^2=1$. Thus no such global holomorphic square root exists. Going once around zero changes the sign of the locally chosen root.

Let $F=e^g$. It is holomorphic and nowhere zero on a punctured neighbourhood of $a$, and the chain rule givesIf $F$ had a pole of order $n\ge 1$, part (c) would imply ${\mathrm{Res}}_a(g^{\prime })=-n$. But the derivative of a holomorphic function on a punctured disc has residue zero. Indeed, write its Laurent seriesTermwise differentiation is valid on every smaller closed annulus. The only possible $(z-a{)}^{-1}$ term comes from $k=0$ and has coefficient zero. Thusa contradiction. Therefore $\boxed{e^g\text{ cannot have a pole at }a}$. This proves an exponential has no isolated pole, including when $g$ has an essential singularity; no choice of a holomorphic logarithm on the entire punctured disc was assumed.

At $z_0$, complex differentiability givesWriting $f^{\prime }(z_0)=\lambda e^{i\alpha }$, the assumption $f^{\prime }(z_0)\ne 0$ gives $\lambda >0$. Its real derivative is therefore the similarity $h\mapsto \lambda e^{i\alpha }h$, a rotation through $\alpha$ followed by scaling by $\lambda$.

More explicitly, for two differentiable curves ${\gamma }_1,{\gamma }_2$ with ${\gamma }_j(0)=z_0$ and nonzero tangent vectors, the chain rule givesTaking the complex argument of this ratio shows that the oriented angle is preserved, while nonvanishing ensures both image tangents remain regular. Thus

Differentiate the family of steady states, keeping $T_i$, $W_C$, $c_C$, $R_1$ and $R_2$ fixed. A quasistatic process uses the temperature responses from part (b):Apply the chain rule to reactor reactivity along that family, weighting each response by its temperature coefficient of reactivity. Collecting terms gives the power coefficient of reactivityIts unit is inverse power. If the thermal resistances, coolant flow or inlet condition varied appreciably with power, their derivatives would contribute additional terms; those changes are excluded by the stipulated constant-parameter model.

Adjoin $\dot{\mu }=0$ and put $w=z+\mu$. The extended centre manifold is a graph $w=H(x,y,\mu )$ tangent to $w=0$ at the origin. The equations becomeHere $ϵ>0$ is fixed. The centre-manifold invariance equation, obtained by the chain rule, isWrite $H=H_2+O({\rho }^3)$, where $\rho =|x|+|y|+|\mu |$ and $H_2$ is a homogeneous quadratic polynomial. To quadratic order the centre-manifold invariance equation isThe terms $\mu x,\mu y,{\mu }^2$ have zero coefficients: their equations are homogeneous, and the operator $1+y{\partial }_x-x{\partial }_y$ is invertible on these quadratic polynomials. For $H_2=a{x}^2+bxy+c{y}^2$, comparison of coefficients givesHence $a=3/5$, $b=-2/5$ and $c=2/5$, givingSubstitution into the $y$ equation givesThe constant in the big O notation may depend on fixed $ϵ$. Thus the equations printed in part (b) are the retained terms of the local Taylor series, rather than an exact identity for the full reduced vector field.

For the Hopf bifurcation classification in part (a), set $\mu =0$ and use $x=r\cos \theta$, $y=-r\sin \theta$. The radial derivative is $\dot{r}=(x\dot{x}+y\dot{y})/r$. Its cubic angular mean isHere $⟨\cdot ⟩$ denotes the mean over one revolution. Removing the oscillatory cubic terms by a near-identity transformation gives this same cubic radial coefficient in the Hopf normal form. The linear radial coefficient is $5ϵ\mu /2$.

The remainder in part (b) involves an unspecified $\xi =\xi (x)$. Its existence for every $x$ does not ensure a differentiable selection. Even if such a selection existed and $f$ had an additional derivative, the product rule and chain rule would giveThe first term cannot be discarded or bounded from the given information. Neither $f^{(n+2)}$ nor ${\xi }^{\prime }$ is supplied by the original theorem.

For an explicit demonstration, interpolate the cubic polynomial $f(x)=x^3$ at zero and one. Then $p_1(x)=x$, andDifferentiating correctly givesTreating the remainder coefficient $x+1$ as a constant would omit $x(x-1)$ and give a wrong answer at every interior point. An existential intermediate point cannot be frozen when differentiating. Part (i) instead applies Rolle's theorem directly to the polynomial interpolation error.

Put $J(x)=DR(x)$, with one residual gradient per row. The chain rule givesDifferentiating each product givesThe approximation $B_n=J(x_n{)}^{T}J(x_n)$ defines the Gauss-Newton method. It is symmetric, and full column rank gives $v^T{B}_{n}v=\Vert J(x_n)v{\Vert }_2^2>0$ for $v\ne 0$, so the step is well-defined.

If $x_n\to x_{\ast }$, $R(x_{\ast })=0$ and $J(x_{\ast })$ has full column rank, then $\nabla f(x_{\ast })=0$ and $H_{\ast }={\nabla }^{2}f(x_{\ast })=J(x_{\ast }{)}^{T}J(x_{\ast })$ is positive definite. Continuity of $J$ gives $B_n\to H_{\ast }$. ThereforeThe Dennis-Moré condition applies, proving $\boxed{x_n\to x_{\ast }\text{ q-superlinearly}}$, or finite termination. The zero-residual hypothesis is what makes the discarded second-derivative term vanish at the limit.

The Jacobian matrix of $F_{\mu }$ in $(x,y)$ isThe top-left block is the Hessian matrix of $f-y^{T}c$ with $y$ held fixed. In the second block, differentiating $C(x)y$ in $x$ gives $YA$ and in $y$ gives $C$. The explicit derivative with respect to $\mu$ is $(0,-e{)}^T$. Differentiate $F_{\mu }(v(\mu ))=0$ by the chain rule:All blocks are evaluated on the current central path point.

By the chain rule, $y^{\prime \prime }=f_x+f_{y}f$ along the exact solution. Applying Taylor's theorem to the exact increment and the two arguments of $f$ givesThe assumed differentiability makes the remainders uniform on $0\le x\le 1-h$. Subtracting gives the consistency errorHere $O$ denotes Big O notation. The general order of a numerical method is therefore one for $\beta \ne 1/2$ and at least two for $\beta =1/2$. For example, on $y^{\prime }=y$ with unit initial value, $y_N=(1+h+\beta h^2{)}^{1/h}$. Expanding its logarithm gives $y_N=e[1+(\beta -1/2)h+O(h^2)]$, so the first-order term is actually nonzero when $\beta \ne 1/2$.

To control accumulation, the mean value theorem and the bound on $f_y$ givePut $K_h=L(1+\beta hL)$, $q_h=1+h{K}_h$ and $e_n=y(x_n)-y_n$. The exact and numerical updates yield $|e_{n+1}|\le q_h|e_n|+h|T_n|$, with $e_0=0$. Iterating this recurrence, as in the Lipschitz one-step error bound, givesBecause $K_h\le L(1+L)$ for $h\le 1$, the prefactor is bounded independently of the mesh. When $\beta =1/2$, the consistency error is $O(h^2)$, hence the maximum global error is $O(h^2)$. This proves at least second-order numerical convergence of the explicit midpoint method.

The chain rule givesThusDifferentiation with respect to a contravariant coordinate has a lower transformation index. In inertial Cartesian coordinates in Minkowski spacetime, no connection term is needed for this coordinate transformation of a derivative acting on a scalar; Lorentz transformations have constant Jacobians.

Set $q=y-x$ and $z_t=x+tq$. The definition of pushforward measure gives the confinement energy $\mathcal{V}[{\mu }_t]=\int V(z_t)d\Pi$. The chain rule and differentiation under the integral sign, whenever justified, giveFor example, bounded Hessian matrix is sufficient: it gives $|\nabla V(z)|\le C(1+|z|)$, and the first two differentiated integrands are bounded by $C^{\prime }(1+|x{|}^2+|y{|}^2)$. The dominated convergence theorem applies because $\Pi$ has finite second absolute moments. Compactly supported endpoints also suffice because all interpolating points lie in one compact set.

The printed growth assumption alone does not ensure a classical second derivative. Here is a counterexample within the stated assumptions. On the line take the bounded nonnegative smooth functionand the probability measures $\mu ={\sum }_n{p}_n{\delta }_{n-1/2}$, $\nu ={\sum }_n{p}_n{\delta }_{n+1/2}$. Both have finite second absolute moments. Their translation transport plan is optimal: its cost is one, while every coupling has cost at least the square of its mean displacement, also one. At $t_0=1/2$, $z_{t_0}=n$, so $F(t)=\mathcal{V}[{\mu }_t]$ satisfies $F(t_0)=0$. Each summand is nonnegative, and the Fatou lemma givesThus the requested classical identity requires an additional integrability assumption on the Hessian matrix; quadratic growth controls the energy but not its second derivative.

For a confinement potential $V$ that is a $\lambda$-strongly convex function, the derivative formula, where valid, yieldsThe underlying displacement convexity conclusion is valid under the printed assumptions without that extra differentiation hypothesis. Apply convexity of $V(x)-\lambda |x{|}^2/2$ pointwise and integrate:The same argument on every subinterval makes $t\mapsto \mathcal{V}[{\mu }_t]-\lambda D^2{t}^2/2$ convex, so the second derivative lower bound also holds in the sense of distributions.

Use an independent copy $(x^{\prime },y^{\prime })$ of $(x,y)$ under the product measure $\Pi \otimes \Pi$. Write $q=y-x$, $q^{\prime }=y^{\prime }-x^{\prime }$, andSince the pushforward measure of $\Pi \otimes \Pi$ under $(z_t,z_t^{\prime })$ is ${\mu }_t\otimes {\mu }_t$, the interaction energy isTwo applications of the chain rule giveThe factor $1/2$ is retained. This classical identity requires differentiation under the integral sign to be justified, just as in part (b). Bounded Hessian matrix of $W$ is sufficient, since $|q-q^{\prime }{|}^2\le 2|q{|}^2+2|q^{\prime }{|}^2$ is integrable; compactly supported endpoints are another sufficient case. The printed quadratic-growth condition alone does not provide this domination.

The failure can occur for the interaction energy too. Take the even interaction potential $W(x)=1-\cos (2\pi x^3)$ and replace the endpoint probability measures in part (b) by $\frac{1}{2}{\delta }_0+\frac{1}{2}\mu$ and $\frac{1}{2}{\delta }_0+\frac{1}{2}\nu$. Keep the origin fixed and translate every other atom by one. This coupling is increasing on the line, hence is optimal: its assigned values extend to a nondecreasing transport map, which is the gradient of a convex function. Differences between translated atoms remain integers, so their mutual interaction vanishes. Only interactions with the origin contribute, giving $\mathcal{W}[{\mu }_t]=F(t)/4$, with $F$ from part (b). It therefore also lacks a finite classical second derivative at $t=1/2$.

For a candidate curve $\rho$, define the velocity field by the convolutionIt is well defined with a finite first absolute moment, because $|K(x-z)|\le L(|x|+|z|)$. Equality of mixed partial derivatives givesThe Hessian matrix bound justifies moving a spatial derivative through this convolution. For smooth densities, the transport equation consequently becomes the continuity equationThis form defines the equation for singular probability measures as well. With initial probability measure ${\rho }_0$, a weak solution is a curve in $C([0,\infty ),{\mathcal{P}}_1)$ such that every $\phi \in C_c^1([0,\infty )\times {\mathbb{R}}^2)$ satisfiesUsing smooth test functions gives the same definition by approximation on compact sets. Continuity in the Wasserstein distance bounds first absolute moments on compact time intervals, so the displayed integrals exist.

For the empirical measure, the chain rule and the particle equations giveIntegrating in time, with zero terminal contribution because $\phi$ has compact support, gives precisely the weak identity and the correct initial empirical measure. Thus ${\mu }^N$ is a global weak solution. The self-interaction term vanishes since $K(0)=0$.