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Differentiability at $x\in {\mathbb{R}}^2$ means that there is a linear map $Df{|}_x:{\mathbb{R}}^2\to \mathbb{R}$ such thatThe partial derivatives are $D_{i}f(x)=Df{|}_x(e_i)$ when the derivative exists; independently, they may be defined by the corresponding one-variable difference quotients.

Suppose the partial derivatives exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points ${\xi }_h,{\eta }_h\to x$ such thatContinuity of the partials makes the remainder after subtractingequal to $o(\Vert h\Vert )$. Hence $f$ is differentiable and this is its derivative.

For the given function, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we haveAt the origin both partial derivatives are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a limit. Neverthelessso $f$ is differentiable at the origin with derivative zero. This illustrates that existence of partial derivatives does not imply their continuity.

The final assertion is false. DefineAgain $|g|\le r^2$, so $g$ is differentiable at the origin and it is smooth elsewhere. But along the $x$-axis,is unbounded near zero, and similarly $D_{2}g(0,y)$ is unbounded. Thus both partial derivatives can be unbounded in every neighbourhood of a point even when the function is differentiable everywhere.

Fix $y\in V$. The map $x\mapsto \theta (x,y)$ is a linear functional, so the finite-dimensional Riesz representation theorem gives a unique vector $\beta (y)$ such thatFor scalars $\lambda ,\mu$,Uniqueness givesso $\beta$ is linear.

For $\alpha \in \mathrm{End}(V)$, apply this result toIt produces a unique linear map ${\alpha }^{\ast }$ satisfyingwhich proves existence and uniqueness of the adjoint.

The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume givesThe free surface and the outlet are both at atmospheric pressure, and their vertical separation is $H+h_0-y$. Applying the Bernoulli equation between them, while retaining the small free-surface speed, givesThereforeThe surface reaches the upper tube end when $y=h_0$. Hence the draining time of a uniform tank through a siphon is

Set $u(r,\theta )=R(r)\Theta (\theta )$. Periodicity and separation of variables give the angular modes $1$, $\cos n\theta$, and $\sin n\theta$. The radial equation iswhose solutions are $r^{\pm n}$ for $n\ge 1$; the zero mode has solutions $1$ and $\log r$. Thus a general real harmonic function on the annulus is

The boundary data contain only the $n=2$ cosine mode, so writeThe condition at $r=a$ gives $B=-A{a}^4$, and the condition at $r=b$ fixes $A$. The Dirichlet problem on an annulus for one Fourier mode therefore has solution

For a general Lagrangian,Along an Euler-Lagrange trajectory,so energy is conserved when $L$ has no explicit time dependence.

For the given Lagrangian, write $v=|\dot{x}|$. At points where $v\ne 0$,The Euler-Lagrange equation is thereforeIn expanded form its left-hand side isThe momentum has fixed magnitude,so this is constant without needing to solve the equation of motion. The Hamiltonian is

The singular Legendre transform of a degree-one velocity Lagrangian applies here: $p$ determines only the direction of $\dot{x}$, not its magnitude, and the velocity Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the Lagrangian from this Hamiltonian.

The Cauchy integral theorem states that if $f$ is holomorphic on a simply connected domain, then its integral around every closed piecewise smooth contour in that domain is zero.

Put $q=b-k$. Completing the square givesAfter the change of variable $z=x-iq/(2a)$, the integral runs along the horizontal line $\mathrm{Im}z=-q/(2a)$. The integrand $e^{-a{z}^2}$ is entire. Apply Cauchy's theorem to a rectangle joining this line to the real axis; the two vertical integrals tend to zero as their real parts tend to $\pm \infty$, because $a>0$. The contour may therefore be shifted to the real line. Hence the Fourier transform of a Gaussian gives

Use the global parametrizationIts tangent vectors areandThus $X$ is a regular parametrization and the graph is a smooth surface.

The first fundamental form isChoosing the upward orientation, the Gauss map is