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The sufficiency claim requires a domain qualification. For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, or more generally on a simply connected open domain, the equivalence is valid. On an arbitrary domain, vanishing curl is necessary but may fail to be sufficient.

Necessity. Suppose every closed-curve line integral vanishes. Fix a point $p$ in the open domain and a unit vector $a$. A sufficiently small planar disk $D_{\epsilon }$ of radius $\epsilon$, centred at $p$ and normal to $a$, lies in the domain. Stokes theorem givesDivide by $\pi {\epsilon }^2$ and let $\epsilon \downarrow 0$. By continuity, the average tends to $(\nabla \times \mathbf{F})(p)\cdot a$. Taking the three coordinate directions proves $\boxed{\nabla \times \mathbf{F}=0}$ everywhere.

Sufficiency on ${\mathbb{R}}^3$. A direct proof constructs a potential of a conservative vector field. PutWrite $F_j$ for the components of $\mathbf{F}$. Vanishing curl says ${\partial }_i{F}_j={\partial }_j{F}_i$. Differentiation under the integral, followed by the chain rule, givesThis is the radial integral potential for a curl-free field. The fundamental theorem for line integrals now gives zero on every closed curve. On a simply connected open domain the analogous argument uses path integrals: every loop can be contracted within the domain, and applying Stokes theorem to a piecewise smooth contraction makes its circulation zero. Equivalently, whenever a particular loop bounds an oriented spanning surface entirely in the domain, Stokes theorem immediately gives zero for that loop.

Why the domain matters. On $U={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, the smooth vector fieldhas zero curl, but its line integral around the unit circle in $z=0$ is $2\pi$, as in ii. No spanning surface for this loop can lie entirely in $U$. The literal claim without a condition on the domain therefore has this counterexample.

Put $h(s)=\sin x(s)-\cos y(s)$. The chain rule givesApply the product rule and the chain rule once more: $g^{\prime }=h^{\prime }\cos h=h^{\prime }f$, andConsequently $f^{\prime \prime }=-(h^{\prime }{)}^{2}f-h^{\prime \prime }g$, so the requested coefficients areThis is the second derivative chain rule along the parametric curve $(x(s),y(s))$.

The middle component suggests a scalar potential with $-{\varphi }_y=2y{\cosh }^2(xz)$, so $\varphi =-y^2{\cosh }^2(xz)+h(x,z)$. The chain rule and $2\sinh t\cosh t=\sinh (2t)$ giveMatching the other two components requires $h_x=h_z=0$. ThereforeIts negative gradient is the given vector field in all three components; the additive constant has no effect on that gradient.

The fundamental theorem of calculus and part (a) giveThe difference is constant in $\theta$. Since $I(0,k)=E(0,k)=0$, the constant is zero, provingThis is simply the chain rule integrated, with $E$ itself as the antiderivative of the square-root factor.

For this homogeneous first-order differential equation, set $v=y/x$, with $x\ne 0$. The product rule gives $y^{\prime }=v+x{v}^{\prime }$, and substitution simplifies the equation toThe useful further substitution is $w=e^{-v}-v^2$. Its chain rule derivative satisfiesThus $(x^{2}w{)}^{\prime }=0$. Replacing $v$ and $w$ gives the implicit solutionNo division by $w$ was needed, so the level $C=0$ is included. A direct verification also identifies the integrating factor for a differential one-form:Setting this exact differential to zero recovers precisely the original equation after multiplication by the nonzero factor $e^{y/x}$. The coefficient $1+2v{e}^v$ never vanishes for real $v$, since $v{e}^v\ge -1/e$ implies $1+2v{e}^v\ge 1-2/e>0$. Hence these level sets give local function graphs $y(x)$ everywhere with $x\ne 0$.

Write $y(\theta )=Y(x)$ with $x=\cos \theta$. By the chain rule,The terms involving $Y_x$ cancel in the original ordinary differential equation, leavingOn each interval where $\sin \theta \ne 0$, the transformed linear ordinary differential equation is $Y_{xx}+2Y=0$. Its characteristic equation has roots $\pm i\sqrt{2}$, yieldingDirect substitution shows these expressions also extend smoothly through $\theta =k\pi$ and satisfy the original equation there. The two constants describe the general solution on any interval where the coordinate substitution is invertible.

For $F(x)=e^{e^x}$, the chain rule gives $F^{\prime }=e^x{e}^{e^x}$ and $F^{\prime \prime }=(e^x+e^{2x})e^{e^x}$. Thus $F(0)=e$, $F^{\prime }(0)=e$ and $F^{\prime \prime }(0)=2e$, yielding

Away from the origin, the chain rule gives $r_x=x/r$ and $r_y=y/r$. Differentiating once more yields the two-dimensional case of the Hessian of the Euclidean norm:These partial derivatives are smooth wherever $r>0$, so Clairaut's theorem gives $\boxed{r_{yx}=r_{xy}=-xy/r^3}$. None of these formulas supplies a derivative at the origin: even the restriction $r(x,0)=|x|$ has no two-sided derivative there.

The initial value is positive, so set $z=y^{-2}$. The chain rule gives $z^{\prime }=-2y^{-3}{y}^{\prime }=-2(y^{-2}-1)$, making $z^{\prime }+2z=2$. Its solution is $z=1+C{e}^{-2x}$. Since $y(0)=1/2$, $z(0)=4$ and $C=3$. Choosing the positive branch givesIt satisfies the initial condition and approaches the stable positive equilibrium one as $x\to \infty$.

For $x=e^z>0$, let $Y(z)=y(e^z)$. The chain rule givesMultiplying each original Cauchy-Euler equation by $x^2$ therefore givesandThe exponential change of variable has made both equations have constant coefficients.

For $0<\theta <\pi$, the chain rule for a parametric curve givesAt $\theta =0$ the quotient of parameter derivatives is formally $0/0$, but the simplified expression has limit $16$. Thus the limiting tangent at $(x,y)=(1,1)$ has gradient 16. This is an endpoint limit of the given open parameter interval, or the ordinary tangent of the smooth polynomial extension $y=8x^4-8x^2+1$.

The product rule and chain rule first giveDifferentiating the first expression with respect to $y$ yields $e^{x^{2}y}(2y+7x^2{y}^2+2x^4{y}^3)$; differentiating the second with respect to $x$ gives exactly the same polynomial factor. ThusThis also follows from symmetry of second derivatives, since all derivatives of this function are continuous everywhere.

The chain rule gives $f{f}^{\prime }=\frac{1}{2}(f^2{)}^{\prime }$. Both $f(0)$ and ${\lim }_{x\to \infty }f(x)$ vanish, so the fundamental theorem of calculus givesThe positive contribution while $f$ rises is cancelled by the negative contribution while it falls.

For fixed $x\ne 0$, the chain rule and the derivative of the hyperbolic secant giveThus the stated antiderivative holds, with its integration constant independent of $y$ but allowed to depend on $x$. At $x=0$ the integrand is zero and the antiderivative is simply constant; the displayed formula with $1/x$ is meant for nonzero $x$.

Work with $t>0$ and write $u=t^{-1/2}{e}^{-x^2/(4t)}$. The chain rule givesThus $u_{xx}=u_t$: it is a constant multiple of the one-dimensional Gaussian heat kernel.

For the modified equation, put $f=g(t)u$. Substitution gives $g{u}_{xx}=g^{\prime }u+g{u}_t-gu/t$. The heat-equation terms cancel, and $u>0$ permits division to obtain $\boxed{g^{\prime }=g/t}$. Solving this separable differential equation givesThe zero solution is included by $C=0$. This is the scalar integrating factor for a reaction-diffusion equation with reaction coefficient $1/t$, because the modified equation is $f_t=f_{xx}+f/t$. Its sign produces growth. Indeed ${\int }_{\mathbb{R}}udx=2\sqrt{\pi }$ is constant, while ${\int }_{\mathbb{R}}fdx=2C\sqrt{\pi }t$ increases linearly for $C>0$; neither profile is being assigned unit normalization here.

At fixed entropy, the total differential in coordinates $(T,p)$ is $0=S_T{|}_{p}dT+S_p{|}_{T}dp$. Dividing by $dT$ along that path gives $S_T{|}_p=-S_p{|}_T{p}_T{|}_S$. The same argument in coordinates $(T,V)$ gives $S_T{|}_V=-S_V{|}_T{V}_T{|}_S$. These are the two required intermediate identities.

To prove the final result without cancelling a possibly vanishing thermal-expansion derivative, write $a=S_T{|}_V$, $b=S_V{|}_T$, $c=p_T{|}_V$, $d=p_V{|}_T$. In a regular thermodynamic region with $a,d\ne 0$, fixed pressure gives $V_T{|}_p=-c/d$, while fixed entropy gives $T_V{|}_S=-b/a$. The chain rule then gives $S_T{|}_p=a-bc/d$ and $p_V{|}_S=d-bc/a$. Both cross-products equal $ad-bc$, provingThe heat-capacity and compressibility ratio identity rewrites this as $C_p/C_V={\kappa }_T/{\kappa }_S$: keeping entropy fixed rather than temperature fixed changes the mechanical stiffness by the same factor as the two thermal responses.

Differentiate the first-order chain rule again, including derivatives of the coordinate coefficients. The second derivatives under a two-variable coordinate change areandThe last two terms account for derivatives of the coordinate functions; they vanish for an affine map. Smoothness justifies equality of the two mixed partial derivatives.

For real $x$, the angle $\arctan x$ lies in $(-\pi /2,\pi /2)$ and has sine $x/\sqrt{1+x^2}$. Thus the principal arcsine in the expression equals the arctangent of $x$. Its derivative isThe chain rule gives the same result directly, using $1-x^2/(1+x^2)=1/(1+x^2)$ with a positive square root.

The chain rule for the cosine gives

On this parametric curve, $x=1$ permits $t=0$ or $t=-2$, but only $t=0$ also gives $y=0$. The chain rule gives the slopeAt the required point, $t=0$, so