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At fixed $z$, differentiating $2x^2+y^2=z^2$ gives $(\partial x/\partial y{)}_z=-y/(2x)$. Also, $x^2=(z^2+u^2)/3$ gives $(\partial x/\partial u{)}_z=u/(3x)$. At fixed $y$, eliminate $x^2$ to write $u^2=(z^2-3y^2)/2$, giving $(\partial u/\partial z{)}_y=z/(2u)$. Finally, the previous part gave $(\partial y/\partial z{)}_u=z/(3y)$. ThusThe held-fixed subscripts matter: substituting a reciprocal partial derivative while silently changing what is held fixed would not preserve these cancellations.

For $F(x,y)=e^{xy}+y\ln x-x^2$, the real domain has $x>0$. Its partial derivatives are $F_x=y{e}^{xy}+y/x-2x$ and $F_y=x{e}^{xy}+\ln x$. At $(1,0)$ they equal $-2$ and $1$, respectively. Thus implicit differentiation gives $y^{\prime }=2$, and the tangent line is

The coordinate partial derivatives are $x_s=-x$, $y_s=-y$, $x_t=y$, and $y_t=-x$. Apply the multivariable chain rule to $v(s,t)=u(x(s,t),y(s,t))$:All partial derivatives of $u$ are evaluated at $(e^{-s}\sin t,e^{-s}\cos t)$. This is a logarithmic polar coordinate change; its Jacobian determinant is $x^2+y^2=e^{-2s}>0$, so it is locally invertible away from the origin. A full angular turn repeats the same point, as usual for polar coordinates.

Set both first partial derivatives equal to zero:Thus $x=0,\pm 1/\sqrt{2}$ and $y=0,\pm \sqrt{2}$, independently, giving exactly nine stationary points. The Hessian matrix is diagonal everywhere:Its diagonal entries are its eigenvalues, so the second-derivative test immediately classifies the points:

In particular,All nine stationary points are nondegenerate, so no higher-order test is needed. Neither type of local extremum is global: $f(x,0)=x^4-x^2\to +\infty$ as $|x|\to \infty$, whereas $f(0,y)=-y^4+4y^2\to -\infty$ as $|y|\to \infty$.

To organize the contour plot, complete the squares:The level set through all four outer saddle points of a scalar function factors exactly:It is therefore the union of the circle $x^2+y^2=5/2$ and the hyperbola $y^2-x^2=3/2$, which cross at those four saddle points of a scalar function. Near the origin, $f=-x^2+4y^2+O((x^2+y^2{)}^2)$, so the zero contours have tangents $y=\pm x/2$. These two exact structural features, together with the nested contours around the four local extrema, determine the requested sketch. They are an example of a separable quartic landscape with nine stationary points.

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.

No scalar potential exists on the full three-dimensional domain. A sufficiently smooth gradient field has zero curl, since mixed partial derivatives commute. The calculated curl is not identically zero. The fact that it vanishes at some individual points does not make the whole field a gradient field.

Integrating the first component and checking the remaining partial derivatives givesThis explicit scalar potential proves that the field is conservative.

Differentiate the rewritten expression. Its new kernel vanishes at the moving upper endpoint, and its partial derivative is an integrable inverse square root. The derivative of an Abel integral isFor a rigorous route, rewrite $2{\int }_0^x{h}^{\prime }(t)\sqrt{x-t}dt$ as ${\int }_0^x{\int }_0^s{h}^{\prime }(t)/\sqrt{s-t}dtds$ using Fubini's theorem, and apply the fundamental theorem of calculus to the continuous inner integral. The assumption $h(0)\ne 0$ implies $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$, so the extension $g(0)=0$ has no finite right derivative at zero.

The gradient is $(f_x,f_y)=(2-2xy,1-x^2)$. Both partial derivatives vanish at $(1,1)$, so it is a stationary point.

The divergence is the sum of the three diagonal partial derivatives:

Each component depends only on its own coordinate, so all cross partial derivatives vanish. Therefore the curl isThis also follows from the identity that the curl of a gradient vanishes for a twice continuously differentiable potential energy.

If $udx+vdy=dF$ with continuous second partial derivatives of $F$, then $u=F_x$ and $v=F_y$. The symmetry of second derivatives gives the necessary condition for an exact differential form:This condition is also locally sufficient on a small rectangle for continuously differentiable coefficients; global sufficiency additionally needs suitable topology of the domain.

Holding $y$ fixed, the chain rule giveswhere the local change of variable from $x$ to $u$ is valid.

For the internal energy $U(S,V)$, its exact differential identifies $T=U_S$ and $p=-U_V$. The Maxwell relation from equality of mixed partial derivatives is $T_V{|}_S=-p_S{|}_V$. At fixed volume, the constant-volume heat capacity satisfiesso $T_S{|}_V=T/C_V$, assuming nonzero $C_V$ and a locally invertible relation. Applying the chain rule to $p(T,V)$ gives the isentropic temperature-volume derivativeFor the stated one-mole ideal gas, the equation of state gives $p_T{|}_V=R/V$ and $C_V=3R/2$, so the right side is $-2T/(3V)$. Along an isentropic process, $p\propto V^{-5/3}$ and $T=pV/R\propto V^{-2/3}$, giving the same left side:

Differentiate the equation for $f$ with respect to $x$, commuting mixed partial derivatives for a sufficiently smooth solution. With $y=f_x$ this givesThis is the viscous Burgers equation when $k>0$.

The denominator $D=x^2+xy+y^2=(x+y/2{)}^2+3y^2/4$ is positive away from the origin. The partial derivatives areThey both vanish exactly onThese are lines of stationary points, with values $1/3$ and $-1$, respectively.

The proposed function has partial derivatives $u_t=1$ and $u_{xx}=2\alpha$. Substitution in the heat equation requires $1=\alpha$, henceThe coefficient $\beta$ remains arbitrary.

For an exact differential $Pdx+Qdy$, equality of mixed partial derivatives requires $P_y=Q_x$. Here this givesOn an open two-dimensional domain the coefficient relation is $x{\mu }^{\prime }+3\mu =0$, so $\mu =C{x}^{-3}$ on a connected interval avoiding zero. The normalization gives $C=1$ on the positive interval through $x=1$. Integrating $f_x=2x^{-3}\sin y$ and checking $f_y$ yieldsThere is no smooth extension of this normalized integrating factor across $x=0$; an independently specified factor on $x<0$ is not fixed by data at one.

Let $a=f(b,c)$ and work where both $f_b$ and $f_c$ are nonzero, so the required inverse partial derivatives exist. At fixed $a$, the total differential gives $0=f_{b}db+f_{c}dc$, hence $(\partial b/\partial c{)}_a=-f_c/f_b$. At fixed $b$, $(\partial c/\partial a{)}_b=1/f_c$. Their product with $f_b$ is $-1$: this is the cyclic partial derivative identity.

For explicit verification in this example, put $P=b^4-6b^2{c}^2+c^4$ and $H=(b^2+c^2{)}^3$. Direct differentiation givesThereforeThe identity is local to regular branches: for these explicit formulas require $bcP\ne 0$. At singular points one cannot justify cancellation of nonexistent inverse derivatives merely by writing a formal product.

Assume $c\ne 0$, so $(\xi ,\eta )=(x-ct,x+ct)$ is an invertible coordinate change. The chain rule givesFor a twice continuously differentiable solution the mixed partial derivatives commute. Squaring the two operators therefore givesThus the one-dimensional wave equation becomes

If $\psi =f(\xi )+g(\eta )$ with twice differentiable $f,g$, its mixed partial derivative vanishes because each summand depends on only one coordinate. Conversely ${\psi }_{\xi \eta }=0$ means ${\psi }_{\xi }$ is independent of $\eta$; integration in $\xi$ supplies $f(\xi )$ and an integration function $g(\eta )$. Hence, on a characteristic rectangle, the general solution isThese terms travel in opposite directions without changing profile, as in D'Alembert's formula.