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On an interval avoiding $x=0$, divide by $x$ to obtain the first-order linear ordinary differential equationAn integrating factor is $\mu (x)=\exp \left(\int -2/xdx\right)=x^{-2}$. Multiplying through and using the product rule givesBy integration by parts, $\int x\cos xdx=x\sin x+\cos x+C$. ConsequentlyThe formula differentiates back to the original equation. Although division excluded $x=0$, this formula has a smooth extension there with $y(0)=0$, satisfying the original equation at zero as well.

Write $\dot{x}=dx/du$, $\ddot{x}=d^{2}x/d{u}^2$, $A_v=dA/dv$, $A_{vv}=d^{2}A/d{v}^2$, and $y_w=dy/dw$, $y_{ww}=d^{2}y/d{w}^2$. Put $S=A+A^T$. The product rule and chain rule giveHere $w^{\prime },w^{\prime \prime }$ are derivatives with respect to $v$. In the first derivative, $x^{T}A\dot{x}={\dot{x}}^T{A}^{T}x$ because the expression is scalar. Likewise ${\dot{x}}^{T}S\dot{x}=2{\dot{x}}^{T}A\dot{x}$ if desired. This differentiation of a parameter-dependent quadratic form does not assume that $A$ is symmetric. The positive linear term $y^{T}b$ contributes the positive chain-rule terms in the $v$ derivatives; it is independent of $u$.

Repeated use of the product rule and chain rule givesThe denominators are strictly positive on the whole real line, making the signs in the next parts easy to read.

Every positive divisor has a unique prime factorization $p_1^{e_1}\cdots p_k^{e_k}$ with $0\le e_j\le i_j$. Conversely every such exponent choice gives a divisor. The exponent choices are independent, so the product rule givesFor $n=1$ the factorization is empty, and the empty product is one, corresponding to the divisor one.

Put $q(x)=x^{T}Ax$ and $t(x)=x^{T}b$. The product rule givesDifferentiate each term. The derivative of $2tAx$ is $2(Ax)b^T+2tA$, while that of $qb$ is $2b(Ax{)}^T$. HenceThe last two terms are transposes of each other, so this Hessian matrix is symmetric, as expected.

The derivative $D_t={\partial }_t+\{\cdot ,H\}$ is a derivation of the Poisson bracket: $D_t\{f,g\}=\{D_{t}f,g\}+\{f,D_{t}g\}$. The time derivative part obeys the product rule, and the Hamiltonian part obeys it by the Jacobi identity. Therefore explicitly time-dependent conserved quantities also have a conserved Poisson bracket.

On an interval where the substitution is real and $x\ne 0$, a factor formed by exponentiation can remove inverse powers from the coefficients of a linear ordinary differential equation. Forthe substitution $y=x^{n}z$ givesIt becomes a constant-coefficient differential equation exactly when $2n+b=0$, $an+d=0$ and $n(n-1)+bn+e=0$. Derive this by the product rule; checking only the coefficient of $z^{\prime }$ is insufficient.

The product rule gives $v^{\prime }=x{y}^{\prime }+y$ for $v=xy$, reducing the displayed equation to the autonomous separable differential equation $v^{\prime }=F(v)$. On an interval where $F(v)\ne 0$, integrate $\int dv/F(v)=x+C$ and recover $y=v/x$ for $x\ne 0$. Zeros of $F$ give additional constant solutions for $v$ and must be considered before division. A solution for $y$ extends through zero only if the recovered quotient has the necessary finite derivatives and satisfies the undivided original equation there.

For solenoidal velocity and magnetic field, with constant magnetic diffusivity $\eta$, the resistive induction equation impliesThe product rule gives $D_{t}P=\mathbf{x}\cdot D_t\mathbf{B}+\mathbf{u}\cdot \mathbf{B}$ and $\Delta P=\mathbf{x}\cdot \Delta \mathbf{B}+2\nabla \cdot \mathbf{B}$. Consequently the radial velocity, through $Q$, is the source term for the radial magnetic field.

// Target: fluid-mechanics.bigb

A vector field directed along the radius from a fixed origin, with magnitude depending only on radius. In three-dimensional Euclidean space and away from the origin, $\mathbf{F}=f(r)\mathbf{r}=rf(r)\hat{\mathbf{r}}$. The product rule gives $\nabla \cdot \mathbf{F}=3f+r{f}^{\prime }$, and the curl vanishes because the derivative matrix is symmetric. These identities alone do not establish regularity at the origin.

On an interval with $x\ne 0$, the substitution $y=xv$ and the product rule give the displayed identity. An equation $x{y}^{\prime }-y=g(y/x)$ becomes $v^{\prime }=g(v)/x^2$ and permits separation of variables wherever $g(v)\ne 0$. Roots $g(v_0)=0$ also give solutions $y=v_{0}x$ and must be retained if division by $g$ was used. This substitution works for this displayed class even though its explicit factor $1/x$ means that $y^{\prime }=y/x+g(y/x)/x$ need not be a homogeneous equation of the usual form $y^{\prime }=F(y/x)$.

Where $q(x)\ne 0$, the product rule and chain rule give the displayed real derivative. Its stationary points therefore satisfy $3q+(x-a)q^{\prime }=0$. The denominator uses the square of the real cube root, which stays positive for negative $q$. Points with $q=0$ must be checked in the original function: a simple zero of $q$ with $x\ne a$ creates a vertical tangent, not a stationary point. If the prefactor also vanishes, a cancellation can change differentiability and must be analyzed separately.

For a twice continuously differentiable regular curve describing motion, let $v=\Vert r^{\prime }(t)\Vert >0$, $T=r^{\prime }/v$, and let $N$ be the principal normal vector where the curvature $\kappa$ is positive. Differentiating the velocity vector $r^{\prime }=vT$ givesThe tangential component changes the speed; the normal component changes the direction of the velocity vector. A particle can therefore have nonzero acceleration at constant speed, as in circular motion with constant speed. The formula follows from $dT/dt=(dT/ds)(ds/dt)=\kappa vN$ and the product rule. When $\kappa =0$, the normal contribution vanishes even though the principal normal vector is undefined there. ^[ref]

In dimensionless coordinates, take $r(t)=(t,t^2)$. At $t=1$,The resulting vector components areBoth increasing speed and turning contribute to the same acceleration.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

The proof is a single time derivative. Apply the product rule to each dot product and use Newton's second law:Integrating from $0$ to $T$ gives the endpoint identity; the stated sublinear condition removes the endpoint term only in the infinite-time limit. Equivalently, for the scalar inertia $I=\frac{1}{2}{\sum }_i{m}_i|{\mathbf{r}}_i{|}^2$ used in the pairwise moment-of-inertia identity, $\dot{I}=G$ and $\ddot{I}=2K+{\sum }_i{\mathbf{r}}_i\cdot {\mathbf{F}}_i$. Specifying this factor of one half avoids confusion with another common inertia convention.

In On a Mechanical Theorem applicable to Heat by R. Clausius (1870, pp. 123-124), the author defines stationary motion through bounded positions and velocities, includes planetary orbits and elastic vibrations, and names the average negative half of the force-position sum the “virial”. That historical convention differs from the auxiliary $G(t)$ defined above. His proof on pp. 126-127 keeps an endpoint term, then explains why it vanishes over a complete period or becomes small in a long bounded irregular motion. The same paper separates internal and external forces and includes uniform boundary pressure on pp. 125-126: confinement was already part of the formulation, rather than a reason to omit external stresses.^[ref]