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For a real exponent $p$ and $x>0$, write $x^p=\exp (p\ln x)$. The chain rule and the derivative of the natural logarithm give $d(x^p)/dx=p{x}^{p-1}$. For a nonnegative integer exponent this agrees with polynomial differentiation and extends to all real $x$. Reversing differentiation gives $\int x^{p}dx=x^{p+1}/(p+1)+C$ for $p\ne -1$ on a suitable real branch. The exceptional exponent $p=-1$ gives $\ln |x|+C$ on either half-line.

For $\gamma \ne 0$ and $x>0$, the change of variables $u=x^{\gamma }$ convertsinto $Y_{uu}+Y=0$, where $y(x)=Y(u)$. The chain rule cancels the first-derivative terms. Thus $\cos (x^{\gamma })$ and $\sin (x^{\gamma })$ form a pair of linearly independent solutions, with Wronskian $\gamma x^{\gamma -1}$. This is a variable-speed reparametrization of the harmonic oscillator equation.

A canonical transformation preserves the symplectic form, equivalently the canonical Poisson brackets, and therefore the Hamiltonian form of every Hamiltonian system. Preservation of just one selected flow is weaker. For $H=(q_1^2+p_1^2)/2$ on two coordinate pairs, $Q_1=q_1$, $P_1=p_1$, $Q_2=2q_2$, $P_2=p_2$ preserves both that physical Hamiltonian and its equations, but $\{Q_2,P_2\}=2$. Thus it is not canonical. Universally preserving Hamiltonian form instead forces $DFJ(DF{)}^T=J$ by applying the chain rule to arbitrary Hamiltonian gradients.

For an isotropic hyperelastic material with incompressibility, the principal stresses are $t_i={\lambda }_i\partial W/\partial {\lambda }_i-p$. This follows by evaluating the Cauchy stress tensor formula $T=F({\partial }_{F}W{)}^T-pI$ in principal directions. For the reduced strain energy density $h(\lambda )=W({\lambda }^{-2},\lambda ,\lambda )$, the chain rule and equality of the tangential derivatives give $t_{\theta }-t_r=\lambda h^{\prime }(\lambda )/2$.

If $\mu$ has Poincaré constant $C$ and $\varphi$ is $L$-Lipschitz, then the pushforward measure ${\varphi }_{\ast }\mu$ has Poincaré constant at most $C{L}^2$. Apply the original inequality to $g\circ \varphi$ and use the chain rule.

On $x>0$, the linear partial differential equation $u_{xx}-x^2{u}_{yy}-u_x/x=0$ has characteristic curves $y\pm x^2/2=\text{constant}$. The change of variables $\xi =x^2-2y$, $\eta =x^2+2y$ has Jacobian determinant $8x\ne 0$. The chain rule givesThus its canonical form of a second-order partial differential equation is $U_{\xi \eta }=0$, and its general twice differentiable solution is $u=f(x^2-2y)+g(x^2+2y)$. The first derivative term exactly cancels the derivatives of the coordinate coefficients.

More generally, for positive differentiable $a(x)$, choose $s(x)$ with $s^{\prime }(x)=a(x)$. Then the same cancellation givesThe characteristic coordinates are $s\pm y$. The quadratic example is $a(x)=x$, $s=x^2/2$, with both coordinates multiplied by two.

The chain rule with $y(x)=Y(x^2)$ gives $x{y}^{\prime \prime }-y^{\prime }=4x^3{Y}^{\prime \prime }$ and transforms the equation, away from zero, into $Y^{\prime \prime }+Y=0$. Thus its solution space is spanned by $\cos (x^2)$ and $\sin (x^2)$. Both extend smoothly through the regular singular point at zero. Their even power series allow the independent data $y(0)$ and $y^{\prime \prime }(0)$ to select the two solutions; every smooth solution has $y^{\prime }(0)=0$.

For $t>0$, the change of variables $\tau =1/t$ and $u(t)=v(\tau )/\tau$ converts $t^4{u}^{\prime \prime }+{\lambda }^{2}u=0$ into $v^{\prime \prime }+{\lambda }^{2}v=0$. The chain rule gives $u^{\prime }=v-\tau v^{\prime }$ and $u^{\prime \prime }={\tau }^3{v}^{\prime \prime }$. Thus $u=t[A\cos (\lambda /t)+B\sin (\lambda /t)]$. The prefactor cancels the derivative term introduced by inverting the independent variable.

Related rates use the chain rule to connect time derivatives of quantities constrained by an equation. If $F(x(t),y(t))=0$, differentiating gives $F_x\dot{x}+F_y\dot{y}=0$. For a rectangle with one fixed side $a$ and variable side $b(t)$, its diagonal $d(t)=\sqrt{a^2+b(t{)}^2}$ obeys $\dot{d}=(b/d)\dot{b}$. The equation must be differentiated before inserting an instantaneous value of $b$, since that value does not make the changing side constant in time.

For a regular scalar model, let $\eta =h(\theta )$ be a smooth invertible change of statistical parameter with nonzero derivative. The chain rule makes the score function in the new coordinate equal to the old score function times $d\theta /d\eta$. Therefore the expected Fisher information transforms as $I_{\eta }=I_{\theta }(d\theta /d\eta {)}^2$. Taking its square root gives exactly the absolute change-of-variable factor required for a prior density.

Consequently, computing the Jeffreys prior directly in either coordinate produces the same measure after transformation. Its density is not numerically constant across coordinates, nor must either density be uniform. When the prior distribution is proper, compatible normalization gives identical prior probabilities for corresponding sets; the resulting posterior distributions also transform consistently. For an improper prior the equality concerns measure kernels up to a constant, and posterior propriety still needs checking.

If $h:[\alpha ,\beta ]\to [a,b]$ is an increasing piecewise smooth change of parameter, with endpoints mapped to endpoints, the chain rule gives $(\gamma \circ h{)}^{\prime }=({\gamma }^{\prime }\circ h)h^{\prime }$. Change of variables in the defining line integral proves the displayed identity. A decreasing change of parameter reverses the endpoints and the sign. The result concerns the same traversal with the same multiplicity; retracing a loop additional times changes the integral in general.

A reversing involution $R$ maps a trajectory $x(t)$ to another trajectory $Rx(-t)$. The chain rule gives the displayed condition. In contrast, an ordinary equivariance symmetry satisfies $F(Sx)=DS(x)F(x)$ and maps trajectories with time unchanged. Oddness $F(-x)=-F(x)$ expresses half-turn equivariance, not reversal by the half-turn. Damping usually breaks the mechanical candidate $R(x,y)=(x,-y)$.

For a constant orthogonal matrix $R$, the Laplacian is unchanged by the coordinate transformation $\xi =Rx$. If $u$ is twice continuously differentiable, the chain rule givesbecause ${\sum }_i{R}_{ji}{R}_{ki}={\delta }_{jk}$ by orthogonality of the rows. Here ${\delta }_{jk}$ is the Kronecker delta. In two dimensions, the rotation matrix $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ with $a^2+b^2=1$ makes the mixed partial derivatives cancel explicitly. The same argument works for reflections and translations, so the Laplacian depends on the Euclidean geometry rather than the orientation or origin of the coordinates.

Let $f$ be a twice continuously differentiable function on an open domain stable under positive scaling, with $f(\lambda x)={\lambda }^{n}f(x)$. Apply the chain rule twice to the scalar function $\lambda \mapsto f(\lambda x)$ and set $\lambda =1$. The result isIn two variables, this reads $x^2{f}_{xx}+2xy{f}_{xy}+y^2{f}_{yy}=n(n-1)f$. In matrix notation it is $x^T({\nabla }^{2}f)x=n(n-1)f$, where ${\nabla }^{2}f$ is the Hessian matrix. Equivalently, the radial differential operator $D={\sum }_i{x}_i{\partial }_i$ obeys $D^{2}f=Df+{\sum }_{i,j}{x}_i{x}_j{\partial }_i{\partial }_{j}f$; substituting $Df=nf$ explains the factor $n(n-1)$.

A second-order reaction has overall order of reaction two: a rate law may depend on $[\mathrm{A}{]}^2$, or on $[\mathrm{A}][\mathrm{B}]$ with partial orders of reaction one and one. Doubling all relevant reactant concentrations multiplies the reaction rate by four at fixed conditions. This experimentally determined dependence does not establish molecularity or a unique reaction mechanism. For an elementary reaction obeying the law of mass action, molecularity and overall order of reaction agree; a net chemical equation alone cannot establish that the conversion is elementary.^[ref]

For the single-concentration model, explicitly choose the disappearance rate conventionHere $k_{\mathrm{obs}}$ is the coefficient for loss of A, held fixed at the specified temperature and other conditions. Its subscript labels the chosen observed disappearance rate; it does not imply a pseudo-first-order reaction. Assume a uniformly mixed closed system, fixed volume, and no appreciable reverse or competing chemical reactions. The integrated rate law isThe reciprocal-concentration plot is a straight line with slope $k_{\mathrm{obs}}$ and intercept $1/[\mathrm{A}{]}_0$. A brief derivation uses the chain rule: $d(1/[\mathrm{A}])/dt=-[\mathrm{A}{]}^{-2}d[\mathrm{A}]/dt=k_{\mathrm{obs}}$, then applies the initial condition. This is a hyperbolic decay, rather than the exponential decay of a first-order reaction.^[ref]

As an original prescribed example, choose $[\mathrm{A}{]}_0=0.020\mathrm{mol}{\mathrm{L}}^{-1}$ and $k_{\mathrm{obs}}=0.50\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. At $t=100\mathrm{s}$, the reciprocal-concentration calculation is $50+50=100\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}$, giving $[\mathrm{A}]=0.010\mathrm{mol}{\mathrm{L}}^{-1}$. The initial disappearance rate is $0.50(0.020{)}^2=2.0\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$; after $100\mathrm{s}$ it is $5.0\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Halving the concentration quarters the disappearance rate.

Dimensional analysis checks every formula: $k_{\mathrm{obs}}[\mathrm{A}{]}^2$ has units $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, $k_{\mathrm{obs}}t$ has units $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}$, and $k_{\mathrm{obs}}[\mathrm{A}{]}_{0}t$ is a dimensionless quantity. Changing concentration units changes the numerical rate constant: the same example is $[\mathrm{A}{]}_0=20\mathrm{mol}{\mathrm{m}}^{-3}$, $k_{\mathrm{obs}}=5.0\times 10^{-4}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$, giving $[\mathrm{A}](100\mathrm{s})=10\mathrm{mol}{\mathrm{m}}^{-3}$.

For an elementary reaction $2\mathrm{A}⟶\mathrm{P}$, instead define the stoichiometrically normalized reaction rate $r=k[\mathrm{A}{]}^2$. Two A entities disappear per P formed, soConsequently the reciprocal-concentration slope is $2k$, not $k$. In the prescribed example, the normalized rate constant would be $k=0.25\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$; initially $r=1.0\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ while the disappearance rate is twice that. With $[\mathrm{P}](0)=0$, stoichiometry gives $[\mathrm{P}](100\mathrm{s})=([\mathrm{A}{]}_0-[\mathrm{A}])/2=0.0050\mathrm{mol}{\mathrm{L}}^{-1}$. Both rate constants have the same dimensions, so a units check alone cannot catch their factor-of-two difference. The symbol $r$ denotes the same normalization as $v$ in reaction rate.

The half-life, meaning the time to reach half a specified starting concentration, follows by setting $[\mathrm{A}]=[\mathrm{A}{]}_0/2$:For $r=k[\mathrm{A}{]}^2$ in $2\mathrm{A}⟶\mathrm{P}$, this becomes $t_{1/2}=1/(2k[\mathrm{A}{]}_0)$. Its units are seconds because $k_{\mathrm{obs}}[\mathrm{A}{]}_0$ has units ${\mathrm{s}}^{-1}$. Doubling the starting concentration halves the first half-life; a quoted half-life needs its starting concentration. This differs from a first-order reaction, whose half-life is independent of starting concentration.^[ref]

Within the prescribed run, $[\mathrm{A}]$ reaches $0.010$, $0.0050$, and $0.0025\mathrm{mol}{\mathrm{L}}^{-1}$ at $100$, $300$, and $700\mathrm{s}$. The successive halving intervals are $100$, $200$, and $400\mathrm{s}$: each begins at a smaller concentration. More generally, reaching fraction $f$ of the initial concentration, with $0<f<1$, requires $t=(f^{-1}-1)/(k_{\mathrm{obs}}[\mathrm{A}{]}_0)$, so $90\%$ consumption takes nine initial half-lives, rather than a fixed sequence of equal waiting times.

A straight reciprocal-concentration plot tests this particular single-concentration rate equation. It does not uniquely identify a reaction mechanism, nor does curvature rule out every second-order reaction: two independently changing reactant concentrations require their own integrated rate law. At low concentration, a small measurement error $\delta [\mathrm{A}]$ becomes approximately $-\delta [\mathrm{A}]/[\mathrm{A}{]}^2$ in the reciprocal, magnifying late-time errors. Use the line as a diagnostic; a nonlinear regression of the original concentrations can retain their measurement scale and explicit error model.

For a positive harmonic function on the unit disc, choose a global harmonic conjugate and apply the Carathéodory derivative bound for the right half-plane to $u+iv$. This gives $|\nabla u(0)|\le 2u(0)$ by the Cauchy-Riemann equations. A disk automorphism sending zero to $z$ has derivative modulus $1-|z{|}^2$, so the chain rule gives the displayed bound. Composing a boundary-pole Poisson kernel with a disk automorphism attains equality, proving the coefficient is optimal at every point.

For hidden activations $h_j=\sigma (b_j+{\sum }_r{W}_{jr}{x}_r)$ and class probabilities $p_c={\mathrm{softmax}}_c(a+Vh)$, a one-hot label $t$ has log-likelihood $\ell ={\sum }_c{t}_c\log p_c$. Put ${\delta }_c=t_c-p_c$ and ${\Delta }_j=h_j(1-h_j){\sum }_c{V}_{cj}{\delta }_c$. ThenThese follow from the chain rule and give single-observation stochastic gradient descent updates for the negative log-likelihood.

If the zero set of a differentiable function $g$ is a submanifold, the chain rule always places its tangent space inside $\ker D{g}_a$. Equality follows from the full-rank hypotheses of the regular level set theorem, but may fail for a singular defining function even when the zero set itself is smooth. For $g:\mathbb{R}\to \mathbb{R}$, $g(x)=x^2$, the zero set is the single point $\{0\}$. Its tangent space is zero, while $D{g}_0=0$ has kernel $\mathbb{R}$. One should therefore check the constraint qualification before using the kernel formula.

For $s>0$, the SIR model equations have the first integral of an ordinary differential equationHere $\ln$ is the natural logarithm. A short proof outline is to use the chain rule and substitute the SIR model equations:The SIR first integral gives $i$ directly from $s$, so it describes the curve in a phase portrait even before its time parametrization is known.

Kinetic energy per unit mass. For an acceleration field $\mathbf{a}$, its change along an actual trajectory is $\Delta (|\mathbf{v}{|}^2/2)=\int \mathbf{a}\cdot d\mathbf{r}$, by the chain rule. This differs by a mass factor from the work integral of a physical force.