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For a unit-speed space curve, setThese form the Frenet trihedron. With the signed torsion convention compatible with the question, the Frenet-Serret formulas areFor an oriented surface with unit normal $N$, the signed geodesic curvature is

On the unit sphere $N=\alpha$. Since $\alpha \cdot t=0$, differentiation givesWrite $\alpha =An+Bb$. Then $A=-1/\kappa$, and differentiating this expression and comparing the $n$ coefficient with $\dot{\alpha }=t$ givesThus, wherever the displayed quantities are defined,Finally,which is the stated curvature decomposition for a spherical curve.

Changing variables in the integral equation,Differentiation gives

The stated operators form a Lax pair when the KdV equation is equivalent toIndeed, if $L\varphi =k^2\varphi$, differentiation and this identity showThus ${\partial }_t+A$ preserves every spectral eigenspace, which is the Isospectral Lax equation formulation.

A grand canonical ensemble describes a system exchanging both energy and particles with a reservoir at fixed $T$, volume, and chemical potential $\mu$. For a noninteracting fermion level $r$, occupation is $0$ or $1$, soFactorization over levels gives $\mathcal{Z}={\prod }_r{\mathcal{Z}}_r$, and differentiation with respect to $\beta \mu$ yields the Fermi-Dirac distribution

Since $\nabla f(x_k)=-r_k$, exact line search minimizes $f(x_k+t{r}_k)$. Differentiation givesso

Termwise differentiation givesThe differentiated series has the same radius of convergence $R$.

Fix $a$ and define $g(b)=e^{a+b}{e}^{-b}$ using only the power series. Termwise differentiation gives $(e^z)\prime =e^z$, and the product rule gives $g\prime (b)=0$. Hence $g$ is constant, so $g(b)=g(0)=e^a$. Multiplication by $e^b$ yields $e^{a+b}=e^a{e}^b$.

For $0<x,y<1$, set $a=x^{-1/n}-1$ and $b=y^{-1/n}-1$. Conditional on $S$,Using part (a),Differentiation givesPutting $y=1$ yields $F_{X_1}(x)=x$, so $X_1$ is uniform on $(0,1)$.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$,Differentiation with respect to proper time givesOrthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A^{\prime \mu }=(0,a^{\prime })$. For collinear motion a Lorentz boost givesFor constant positive proper acceleration $a^{\prime }$,This hyperbola has future null asymptote $x=ct-c^2/a^{\prime }$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a^{\prime }$, it lies beyond that acceleration horizon and never intersects the worldline.

For incompressible inviscid flow without body force,Taking curl givesequivalently ${\partial }_t\omega =\nabla \times (u\times \omega )$.

Direct differentiation of the stated field gives $\omega =u$. Hence $u\times \omega =0$, so the vorticity equation gives ${\partial }_t\omega =0$ and the field is steady. FinallyEuler's equation therefore impliesso $H$ is spatially uniform.

The local maximum modulus principle says that if $|f|$ has a local maximum at an interior point of a connected domain, then $f$ is constant. On a small circle about the maximum, the mean-value property andforce equality everywhere. Equality in the triangle inequality makes the boundary values identical; Cauchy's formula then makes $f$ constant locally, and the identity theorem makes it constant on the domain.

Write $f=u+iv$. The hypothesis gives $u-v\le 0$. Thereforewith equality at zero. The local maximum modulus principle makes the exponential constant. Differentiation then gives $f^{\prime }=0$, so $f$ is constant; since $f(0)=0$, it is identically zero.

Conditionally on $X$, the log likelihood differs fromonly by terms independent of $\beta$. Differentiation gives the weighted normal equationsand therefore

The one-particle partition function iswhere $\lambda =(2\pi {\hslash }^2\beta /m{)}^{1/2}$. ThusCanonical differentiation now givesandThereforeThe factor $5/2$ consists of $3/2$ from kinetic energy and $1$ from gravitational potential energy.

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 $x^2{\int }_0^1{y}^4{e}^{xy}dy-12{\int }_0^1{y}^2{e}^{xy}dy=e^x(x-4)$.

Off the slit choose the stated analytic square-root branch. Differentiation gives $f\prime (z)=c^{-1}(1+z/\sqrt{z^2-c^2})$, which does not vanish there, so the map is conformal.

Separation gives angular modes $e^{im\theta }$ and $-[rR\prime ]\prime +m^{2}R/r=\lambda rR$. Frobenius gives $p=m$ and $a_k=-a_{k-2}/[k(k+2m)]$, producing $J_m$; termwise differentiation gives the two stated recurrences. A radial solution is $u={\sum }_n[A_n\cos (c{j}_{0n}t)+B_n\sin (c{j}_{0n}t)]J_0(j_{0n}r)$. Here $B_n=0$ andequivalently $A_n=2[j_{0n}{J}_0\prime (j_{0n}{)}^2{]}^{-1}{\int }_0^{1}r{J}_1(j_{0n}r)dr$ after integration by parts.

With the proton fixed, $[-{\hslash }^2{\nabla }^2/(2m_{\mu })-e^2/(4\pi {\epsilon }_{0}r)]\psi =E\psi$: kinetic energy plus Coulomb attraction, assuming nonrelativistic point particles and neglecting spin, radiation, and proton motion (or replacing $m_{\mu }$ by the reduced mass). Substitution of the $2s$ form gives $E=-{\hslash }^2{\alpha }^2/(8m_{\mu })$ and $\alpha =m_{\mu }{e}^2/(4\pi {\epsilon }_0{\hslash }^2)$. Direct differentiation verifies it. The curve starts positive, crosses at $r=2/\alpha$, then approaches zero from below; ordinary hydrogen has the same shape but the smaller electron mass gives a larger radial scale.

Direct differentiation verifies the free-particle Schrödinger equation, so $V=0$. At $t=0$, normalisation of $A{e}^{-x^2/2}$ gives $A={\pi }^{-1/4}$. Symmetry gives $⟨x⟩=0$, while Gaussian integration givesIts growth shows wave-packet spreading and therefore a nonstationary state.

Differentiation givesequivalently the Laplace-pressure condition

Differentiation gives $u^{\prime \prime }+u=\frac{3}{2}{r}_s{u}^2$. With $u=b^{-1}(v_0+\epsilon v_1+\cdots )$,The symmetry selects $v_0=\sin \varphi$, the straight line at closest distance $b$. Since
$\frac{3}{2}{\sin }^2\varphi =\frac{3}{4}(1-\cos 2\varphi )$,The two asymptotic zeros shift from $0,\pi$ to $-\epsilon ,\pi +\epsilon$, so the deflection is $2\epsilon =2r_s/b=4GM/(b{c}^2)$, the Schwarzschild light deflection. It is tested by stellar deflection near the Sun and gravitational lensing.