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Let ${\mathrm{Skew}}_{2n}$ be the skew-symmetric $2n$ by $2n$ matrices and defineThen $S{p}_n(\mathbb{R})=F^{-1}(J)$, andAt a symplectic $A$, put $H=AX$. The derivative becomes $X^{T}J+JX$. Given skew $K$, choosing $X=-\frac{1}{2}JK$ gives $X^{T}J+JX=K$, so the derivative is surjective.

The preimage theorem makes the level set a submanifold. Sinceits dimension is $2n^2+n$.

Equality of the mixed derivatives givesFor a fundamental matrix solution this is equivalent to the zero-curvature condition

PutAway from $x=x^{\prime }$, this satisfies $(d^2/d{x}^2+k^2)G_k=0$. Its first derivative has jump $2ik$, so, distributionally,Applying $d^2/d{x}^2+k^2$ to the proposed integral equation therefore givesSince $k^2=2mE/{\hslash }^2$, this is precisely the stated Schrödinger equation. The term $e^{ikx}$ is the incident wave and the choice $e^{ik|x-x^{\prime }|}$ gives outgoing waves on both sides of each source point.

The change of variable $x=(y/\lambda {)}^k$ has derivative $kx/y$, soFor fixed $k$ this isThus the sample's sufficient statistic is $T={\sum }_i{Y}_i^k$.

Since $(Y/\lambda {)}^k\sim \mathrm{Exp}(1)$,The log likelihood, up to constants, is $-kn\log \lambda -T/{\lambda }^k$. Differentiating gives the unique maximum

With $\xi =x-ct$, time derivatives become $-c$ times $\xi$ derivatives and $I_{xx}=I^{\prime \prime }$. Henceor equivalently

On the positive real axis, whose path does not cross the left-displaced cut,Since $F_B^{\prime }(z)=1/(1+z^2)=O(|z{|}^{-2})$ near infinity, integrating this derivative along large arcs extends the same expansion uniformly there:Thereforeand hence

If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then some $c\in (a,b)$ has $f\prime (c)=0$. A maximum or minimum is attained; unless $f$ is constant, one extremum lies inside, where the derivative vanishes.

False. The Cantor function has derivative $0$ outside the Cantor set, hence its derivative extends to the Riemann-integrable zero function, while $F(1)-F(0)=1$.

If $f$ has $n+1$ derivatives near $x_0$, then $f(x)={\sum }_{k=0}^n{f}^{(k)}(x_0)(x-x_0{)}^k/k!+f^{(n+1)}(\xi )(x-x_0{)}^{n+1}/(n+1)!$ for some intermediate $\xi$.

$e^x$ is real analytic because $e^{x_0+h}=e^{x_0}{\sum }_{n\ge 0}{h}^n/n!$; it is not a polynomial since all its derivatives are nonzero.

Differentiation under the integral gives $I\prime \prime ={\int }_0^1{y}^4{e}^{xy}dy$. Integrating the derivative of $e^{xy}(x{y}^3-3y^2)$ and then once more yields $x^2{\int }_0^1{y}^4{e}^{xy}dy-12{\int }_0^1{y}^2{e}^{xy}dy=e^x(x-4)$.

Put $T=\pi /\omega$, $t_0=T/2$, and $\Omega =\omega /2$. A solution satisfying both endpoint conditions iswhere $A=-\sqrt{2}I{\omega }^{-1}{e}^{\sqrt{3}\pi /4}$. The derivative jumps by $I$ at $t_0$, as required.

Spatial derivatives vanish for $u=y(t)$, giving $\ddot{y}+\gamma \dot{y}+{\omega }^{2}y=f(t)$.

Using ${\epsilon }_{ijk}{\epsilon }_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$,
$[\nabla \times (F\times G){]}_i={\partial }_j(F_i{G}_j)-{\partial }_j(F_j{G}_i)$; expanding the derivatives gives the stated four terms.

Using $\nabla |x-x\prime {|}^{-1}=-\nabla \prime |x-x\prime {|}^{-1}$, the divergence is a line integral of a total derivative, zero for a closed wire or suitable vanishing endpoints. Taking the curl and moving it under the integral gives the stated Biot–Savart formula.

For fixed $x$ add $K\Pi (t)$ to $p(t)-f(t)$ so the resulting function also vanishes at $t=x$. Repeated Rolle gives a point $\xi$ where its $(n+1)$st derivative vanishes, yielding the formula.

The inverse function theorem says that a $C^1$ map with invertible derivative at $x$ restricts to a $C^1$ diffeomorphism between neighbourhoods of $x$ and $f(x)$. It makes every zero isolated. Compactness of $\bar{\Omega }$, together with separation from zero on the boundary, then makes the zero set finite and interior. Continuity of $Df$ gives the displayed radii. Choose disjoint balls around the zeros and a positive lower bound for $|f|$ on their complement and boundary. A sufficiently small $C^1$ perturbation has no zeros outside, while the quoted local lemma gives exactly one zero in each ball with the same determinant sign. Summing proves $N(g)=N(f)$.