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For $D,\gamma >0$, the displayed function solves the heat equation $u_t=D{u}_{xx}$ and has initial condition $u(x,0)=A{e}^{-\gamma (x-x_0{)}^2}$. Set $s=1+4\gamma Dt$. Direct differentiation givesThese identities prove the equation. The total integral is $A\sqrt{\pi /\gamma }$, independent of time. The real smooth formula is valid for $t>-1/(4\gamma D)$, including a finite interval of negative time for this particular initial profile. For nonzero $A$, its value at $x_0$ diverges as the lower time endpoint is approached. The superposition principle combines finitely many such profiles with different centers or widths.

Since $k=\omega /c(\omega )$, differentiation gives $u=(dk/d\omega {)}^{-1}=c/[1-(\omega /c)dc/d\omega ]$. Using the period $P=2\pi /\omega$ gives the displayed formula. A positive period slope makes the group velocity smaller than the phase velocity on this branch. A slope read from a graph produces an approximate local result, not a globally fitted dispersion law.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

For the first-order ordinary differential equation $y^{\prime }=f(x,y)$ defined on an open subset $D$ of the plane, an isocline of slope $k$ is the level setHere $k$ is a constant. At every point of this isocline, the direction field prescribes a tangent vector proportional to $(1,k)$ for a solution of a differential equation. It does not generally prescribe the tangent line of the isocline itself. The level set can have several components or fail to be a smooth curve, and points outside $D$ are excluded.

For example, take $y^{\prime }=x-y$ on the whole plane. Its isoclines are the parallel straight linesEach line has slope $1$, while the direction field along it has slope $k$. Thus the isocline $y=x$ carries horizontal segments even though the line itself rises. A function graph of a solution of a differential equation follows the segments, rather than following this line. Only the isocline with $k=1$, namely $y=x-1$, is also a solution of a differential equation.^[ref]

The first-order linear differential equation has the exact familyThe initial condition $y(x_0)=y_0$ selects $A=(y_0-x_0+1)e^{x_0}$ and hence a unique solution of a differential equation for every real $x$. The picture shows $y(0)=-1,0,1,2$. For $A=0$ the solution of a differential equation stays on the $k=1$ isocline; for other $A$ it moves through different isoclines. These formulas check the pictured trajectories by direct differentiation, independently of how the direction field was sampled.

The zero-slope isocline $f(x,y)=0$ is called a nullcline. Below the nullcline $y=x$ in this example, $y^{\prime }>0$; above it, $y^{\prime }<0$. For $A>0$, the solution of a differential equation meets it at $x=\log A$, where $y^{\prime }=0$ and $y^{\prime \prime }=1$, a strict local minimum. The nullcline is therefore a useful guide to turning points, although it is not itself a solution of a differential equation. In the equivalent planar autonomous differential equation $\dot{x}=1$, $\dot{y}=x-y$, it is the nullcline for the second component; there is no equilibrium point, because the first component never vanishes.

This makes isoclines useful even when no explicit solution of a differential equation is available: organize a direction field by common slopes, read where function graphs rise or fall, and identify possible turning points on nullclines.

Use local east, north, and up unit vectors, with $x$ eastward, $y$ northward, and positive orientation $\psi$ anticlockwise as viewed from above. Let $\phi$ be signed geographical latitude, positive north, and $\Omega >0$ the magnitude of Earth's angular velocity. In the spherical-Earth approximation,Here $g$ is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with $\ell$, the horizontal restoring acceleration from tension is $-{\omega }_0^2(x,y)$. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equationsThese equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order ${\Omega }^2$; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number $z=x+iy$. Multiplying the second equation by $i$ and adding givesSet $z=e^{-i\alpha t}u$. Direct differentiation cancels the first derivative of $u$ and yieldsThe ${\alpha }^2$ in $\beta$ comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order ${\Omega }^2$. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity $-\alpha$ relative to the building. When $|\alpha |\ll {\omega }_0$, each rapid swing samples an almost fixed long axis; its slow precession isThe sidereal day $T_{\mathrm{sid}}=2\pi /\Omega$ is a rotation relative to distant directions. $T_{\mathrm{p}}$ denotes a full $2\pi$ turn of a continuously followed orientation; an unlabelled swing line repeats after $T_{\mathrm{p}}/2$. At the equator there is no finite precession period in this model. The Coriolis parameter is $f=2\alpha$, so the plane's rate is $-f/2$, not $-f$.

The limiting cases check the sign and scale. At the north pole $\alpha =\Omega$: the swing direction turns clockwise once per sidereal day. At the equator $\alpha =0$: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole $\alpha =-\Omega$: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, the sidereal day is approximately $23.9345\mathrm{h}$. At $45^{\circ }$ north the predicted rate is $-10.6356^{\circ }/\mathrm{h}$ and $T_{\mathrm{p}}\simeq 33.8485\mathrm{h}$; at $45^{\circ }$ south the rate changes sign. At the illustrative geographical latitude $34^{\circ }$ south, the prediction is $+8.4109^{\circ }/\mathrm{h}$. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]

Choose an original simulated model: a primitive cubic lattice with one identical point scatterer per primitive unit cell, edge $a=0.400\mathrm{nm}$, and incident X-ray wavelength $\lambda =0.200\mathrm{nm}$. These are stipulated lengths, not measurements of sodium chloride. Suppose orientation information has already assigned the two observed reflections the Miller indices $(100)$ and $(110)$. The detector deflection is $2\theta$, where $\theta$ is the Bragg angle.

Use the existing interplanar spacing formula and Bragg's law, with higher diffraction orders absorbed into the reflection indices:The simulated $(100)$ reflection has $\sin \theta =1/4$, hence $2\theta =28.9550^{\circ }$. The $(110)$ reflection has $\sin \theta =\sqrt{2}/4$, hence $2\theta =41.4096^{\circ }$. Their squared sines are $0.0625$ and $0.125$, in ratio $1:2$. Substituting the unrounded angles independently givesAgreement tests a proposed indexing and scale together; disagreement would prompt checks of the indexing, wavelength, angular zero, or the assumed unit cell geometry.

Two peak positions alone do not establish the indexing. Assigning the same two angles to $(110)$ and $(200)$ would give $N=2,4$ and $a=0.400\sqrt{2}\simeq 0.565685\mathrm{nm}$ instead. More reflections, orientation information and systematic absences help discriminate such assignments. An unobserved reflection can be inaccessible, weak or extinguished by the atomic basis; absence is not automatically evidence for a larger interplanar spacing.

Absolute scale also requires measurement calibration of the wavelength. To first order, differentiation of the inferred length giveswith angular changes in radians. A stipulated detector-zero error $\delta (2\theta )=+0.100^{\circ }$, with correct wavelength, biases the first inferred edge downward by approximately $0.338\%$ and the second by $0.231\%$. A common angular offset therefore need not cancel when two inferred lengths are compared. These are model sensitivity estimates, not historical instrument uncertainties.

In the quadrant $x,y>0$, a line vortex of circulation $\Gamma$ at $(x,y)$ has three images: circulations $-\Gamma$ at $(-x,y)$ and $(x,-y)$, and $+\Gamma$ at $(-x,-y)$. Their induced velocity isIn polar coordinates, these equations give $\dot{r}=\Gamma \cot (2\theta )/(2\pi r)$ and $\dot{\theta }=-\Gamma /(4\pi r^2)$. Direct differentiation yields $\frac{d}{dt}(r\sin 2\theta )=0$, so the trajectory obeys $r\sin 2\theta =\text{constant}$.

Differentiation preserves the ordering of interlacing roots of polynomials of equal or consecutive polynomial degrees. Iteration gives the result for every derivative for which both polynomials are nonconstant; common roots are allowed by a limiting argument. This is a result about the locations of roots of a polynomial, distinct from the Markov inequality for polynomial derivatives. Theorem A of Dimitar K. Dimitrov's account states the first-derivative result.

Scale the catenoid by its waist radius $a>0$ and choose a unit normal with outward radial component:The hyperbolic cosine controls the profile $r(z)=a\cosh (z/a)$. Direct differentiation yields the first fundamental form and second fundamental formThe coordinate tangent vectors are orthogonal, so the meridional and circumferential principal curvatures areThe opposite principal curvatures cancel at every point, not just at the waist. This is precisely why the catenoid is a minimal surface, while its nonzero second fundamental form and negative Gaussian curvature show that it is not a plane.

Minimizing $\Vert y-X\theta {\Vert }^2$ gives the displayed equations by differentiation. Equivalently, the regression residual $r=y-X\hat{\theta }$ satisfies $X^{T}r=0$: it is orthogonal to every column of the design matrix. If $X$ has full column rank, the minimizer is unique and the fitted values are the orthogonal projection of $y$ onto its column space. A rank-deficient design matrix can give multiple coefficient vectors, while the projected fitted vector is still unique.

For an intercept and one predictor, put $S_{xx}={\sum }_i(x_i-\bar{x}{)}^2$ and $S_{xy}={\sum }_i(x_i-\bar{x})(y_i-\bar{y})$. The equations give $\hat{\beta }=S_{xy}/S_{xx}$ and $\hat{\alpha }=\bar{y}-\hat{\beta }\bar{x}$, provided $S_{xx}>0$. In the five-point example in linear regression, $\bar{x}=2$, $\bar{y}=2.8$, $S_{xx}=10$ and $S_{xy}=10$, giving $\hat{\beta }=1$ and $\hat{\alpha }=0.8$.

The normal equations describe the mathematical optimum, but explicitly forming and inverting $X^{T}X$ can amplify numerical problems. A QR decomposition computes the same least-squares solution through orthogonal coordinates; a singular value decomposition also exposes nearly dependent columns. This connects statistical fitting to numerical linear algebra and the choice of a stable algorithm. ^[ref]

Twice differentiation the sine givesBoth Dirichlet boundary conditions hold when $n$ is an integer. The zero function at $n=0$ is not an eigenfunction, and negative $n$ merely changes an overall sign. ThereforeThe corresponding normalized eigenfunction is $\sqrt{2/a}\sin (n\pi x/a)$.

For distinct positions and positive masses, Newton's law of universal gravitation givesThe proposed homothetic gravitational motion has ${\ddot{\mathbf{x}}}_i=-(2/9)t^{-4/3}{\mathbf{a}}_i$ and pair distances $t^{2/3}|{\mathbf{a}}_i-{\mathbf{a}}_j|$, so cancellation of $t^{-4/3}$ gives the time-independent equationsThese define a Newtonian central configuration with this choice of scale. Fordifferentiation in ${\mathbf{a}}_i$ gives ${\nabla }_{{\mathbf{a}}_i}\Phi =(2/9)m_i{\mathbf{a}}_i-G{\sum }_{j\ne i}{m}_i{m}_j({\mathbf{a}}_i-{\mathbf{a}}_j)/|{\mathbf{a}}_i-{\mathbf{a}}_j{|}^3=0$, exactly the displayed equations. The double sum in the paper counts each pair twice, so it agrees with this expression.

Multiply the configuration equations by $m_i$ and sum. Pair contributions cancel, giving ${\sum }_i{m}_i{\mathbf{a}}_i=0$. Thus the total linear momentum is $\mathbf{P}=(2/3)t^{-1/3}{\sum }_i{m}_i{\mathbf{a}}_i=0$. Each position is parallel to its velocity, so every $m_i{\mathbf{x}}_i\times {\dot{\mathbf{x}}}_i=0$ and the total angular momentum is zero. About any other fixed point $\mathbf{b}$, it is ${\mathbf{L}}_{\mathbf{b}}={\mathbf{L}}_O-\mathbf{b}\times \mathbf{P}=\boxed{0}$.

Here the inverse in the elliptic curve group law is $(x,y)\mapsto (x,-y-1)$. For $P_0+P_1$, the line is $y=2-2x$. Substitution in the Weierstrass equation of an elliptic curve givesThe third intersection is $(3,-4)$, so reflection givesFor the difference, first take $-P_1=(1,-1)$. The line through $P_0$ and $-P_1$ is $y=2-3x$, and now the intersection polynomial is $x(x-1)(x-8)$. Its third point is $(8,-22)$, givingFinally, implicit differentiation gives $(2y+1)y^{\prime }=3x^2-7$. The tangent line at $P_1$ therefore has slope $-4$ and equation $y=4-4x$. Its intersection polynomial factors as $(x-1{)}^2(x-14)$. The double root records the intersection multiplicity of the tangent, and the remaining intersection is $(14,-52)$. Hence

The Markov interlacing lemma says that differentiation preserves interlacing roots of polynomials. More explicitly, if two degree-$d$ real polynomials have ordered real roots satisfying $a_1\le b_1\le a_2\le \cdots \le a_d\le b_d$, their $k$th derivatives have ordered roots $a_j^{(k)},b_j^{(k)}$ satisfyingThe analogous alternating order is preserved for polynomials of consecutive degrees. Common roots are allowed through weak inequalities; strictly interlacing simple roots give strict inequalities. When a derivative is constant, its root list is empty.

Put $V=\mathbb{E}{X}_n^2$. For $c\ge 0$, the process $(X_m+c{)}^2$ is a nonnegative submartingale, by the conditional Jensen inequality. A crossing of $x$ by $X$ implies a crossing of $(x+c{)}^2$ by this squared process. Since $\mathbb{E}{X}_n=\mathbb{E}{X}_0=0$, the Doob maximal inequality givesFor $V>0$, differentiation of the last expression gives $2(cx-V)/(x+c{)}^3$, so its minimum over $c\ge 0$ occurs at $c=V/x$. Substitution proves the one-sided martingale maximal inequalityIf $V=0$, then $X_n=0$ almost surely and $X_m=\mathbb{E}[X_n\mid {\mathcal{F}}_m]=0$ for every $m\le n$, so the same conclusion holds. The constant is sharp: at one step take $X_1=x$ with probability $V/(V+x^2)$ and $X_1=-V/x$ otherwise. This has mean zero, variance $V$, and achieves equality.

Finite second moment implies $\mathbb{E}X<\infty$. For $\lambda$ near any positive ${\lambda }_0$, the derivative of $e^{-\lambda X}$ is bounded in absolute value by $X$, provided the neighbourhood stays in $(0,\infty )$. The dominated convergence theorem therefore permits differentiation under the expectation:For the expansion, put $r(z)=e^{-z}-1+z-z^2/2$. Taylor expansion gives $r(z)/z^2\to 0$ as $z\downarrow 0$. Also, for $z\ge 0$,The inequalities follow by integrating $0\le 1-e^{-u}\le u$ from zero to $z$. Consequently $r(\lambda X)/{\lambda }^2\to 0$ pointwise and its absolute value is bounded by $X^2/2$. Another application of the dominated convergence theorem gives the second-moment expansion of a Laplace transform:No third moment is needed for the remainder estimate.

Use uniform-envelope rejection sampling for a beta distribution. Put $r(x)=x^{a-1}(1-x{)}^{b-1}$ for $0<x<1$ and choose $M=\sup r$. For $a,b>1$, differentiation of $\log r$ gives the mode $(a-1)/(a+b-2)$, soIf $a=1$ or $b=1$, take $M=1$; this also covers $a=b=1$. Draw $X=U_{2j-1}$ and accept it ifOtherwise continue with the next independent pair. The probability of acceptance is ${\int }_0^{1}r(x)dx/M=B(a,b)/M>0$, so the procedure terminates almost surely. Conditional on acceptance, the probability density function is proportional to $r(x)$, hence is exactly the beta distribution of parameters $a,b$. The restriction $a,b\ge 1$ guarantees the bounded uniform envelope.

An integrating factor isMultiplication by this integrating factor turns the first-order linear differential equation into $(e^{x^3}y{)}^{\prime }=x^2{e}^{x^3}$. Integrating from zero to $x$ givesThus the initial value problem has the solutionIndeed, direct differentiation gives $y^{\prime }=-3x^2(a-1/3)e^{-x^3}$, which verifies both the differential equation and the initial condition.

Substitution of the power series, with termwise differentiation, gives the coefficient identityThus $a_0$ and $a_1$ are arbitrary, and all power series solutions of the Hermite differential equation areAn empty product is one. For a nonterminating branch the ratio of consecutive terms in its own parity subsequence is asymptotic to $|x{|}^2/j$, hence tends to zero for every fixed $x$. The ratio test proves convergence everywhere; the resulting power series can be differentiated termwise and indeed solve the differential equation. A terminating branch is already a polynomial.

The boundary condition $W(0)=0$ forces $a_0=0$; the solution is odd. Apart from the zero solution, which works for every $E$, a polynomial occurs exactly when an odd index $d$ satisfies $2d-E=0$. Before that index the odd coefficients are nonzero, and after it they vanish. Thus the odd polynomial solutions of the Hermite differential equation haveEquivalently, comparison of leading polynomial coefficients gives $E=2\mathrm{deg}W$, proving that no other nonzero polynomial is possible. Normalizing by $a_1=1$, the polynomials of degree below six areEach can be multiplied by an arbitrary nonzero constant; these are normalized odd Hermite polynomials.

Let $L=-d^2/d{x}^2+2xd/dx$. The forcing in the final inhomogeneous linear differential equation is precisely the displayed degree-five eigenfunction $P_5$, with $L{P}_5=10P_5$. HenceDirect differentiation gives $LW=x-4x^3/3+4x^5/15$, and $W(0)=0$. Among polynomial solutions this choice is unique: a polynomial in the kernel of $L$ must be constant, and its value at zero eliminates that constant.

For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field, Stokes theorem stateswhere the boundary orientation is induced by $n$ through the right-hand rule. Choose the upward orientation of the elliptic paraboloid. Its band has cylindrical radius $1/2\le r\le 1$, and a parametrization isIts oriented surface element and scalar area element areThe requested sketch shows the open band, not a capped solid.

For the specified vector field, direct differentiation gives $\nabla \times B=(0,0,3x^2+3y^2)$. HenceTo check this by Stokes theorem, parametrize a counterclockwise circle of radius $\rho$ at height ${\rho }^2$ by $r_{\rho }(\phi )=(\rho \cos \phi ,\rho \sin \phi ,{\rho }^2)$. Along it $dz=0$, andUsing ${\int }_0^{2\pi }{\sin }^4\phi d\phi ={\int }_0^{2\pi }{\cos }^4\phi d\phi =3\pi /4$ gives $3\pi {\rho }^4/2$. The induced outer circle is counterclockwise as viewed from above, and the inner circle is clockwise. ThereforeThis verifies the direct surface integral and the Stokes flux through an annular paraboloid. Reversing the chosen surface orientation reverses the final sign.