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For $y\prime \prime +py\prime +qy=0$, the Wronskian satisfies $W\prime =-pW$.

More precisely, let $y_1,y_2$ be solutions of a differential equation for the same homogeneous linear differential equationon an interval $I$ where $p,q$ are continuous functions. For any $x_0\in I$, Abel's identity givesOnly the coefficient of $y^{\prime }$ enters the evolution of the Wronskian. In an unnormalized equation $A(x)y^{\prime \prime }+B(x)y^{\prime }+C(x)y=0$, first use $p=B/A$ on an interval where $A\ne 0$ and the coefficient ratios are continuous functions.^[ref]

The proof uses cancellation. Differentiating the Wronskian by the product rule and substituting the common equation givesMultiplying this first-order linear differential equation by the integrating factor $\mu (x)=\exp ({\int }_{x_0}^{x}p(s)ds)$ gives $(\mu W{)}^{\prime }=0$. Since $\mu (x_0)=1$, the constant is $W(x_0)$, proving Abel's identity. This argument also covers $W(x_0)=0$ without dividing by $W$.

The exponential function is positive for real $p$, so a real Wronskian either vanishes everywhere or has a fixed nonzero sign throughout $I$. In the preceding example, $p=-1$ and $W(0)=3$ give $W(x)=3e^x$ immediately. If $p=0$, every Wronskian of two solutions of a differential equation for that same equation is constant, regardless of $q$. Thus Abel's identity can certify linear independence and find the denominator used in variation of parameters without explicit formulas for both solutions of a differential equation.

The arithmetic derivative on positive integers is defined by $p\prime =1$ for every prime $p$ and the product rule $(ab)\prime =a\prime b+ab\prime$. For $n={\prod }_i{p}_i^{e_i}$, these rules give $n\prime =n{\sum }_i{e}_i/p_i$. In particular, $(p^e)\prime =e{p}^{e-1}$.

For arbitrary real $n\ne 0,1$, work first on an interval where $y>0$. Define exponentiation by $y^r=\exp (r\log y)$, using the real logarithm. The change of variable is differentiable and invertible there, and the chain rule givesMultiplying the Bernoulli differential equation by $(1-n)y^{-n}$ yields the first-order linear differential equationThis is an exact transformation of nonzero solutions of a differential equation on the chosen branch; it is valid far from any reference state and does not discard higher powers as a local linearization would.

Given the initial condition $y(x_0)=y_0>0$, the integrating factor and transformed solution of a differential equation areThe product rule verifies $(\mu z{)}^{\prime }=(1-n)\mu Q$. Recover $y=z^{1/(1-n)}$ only on the connected component of $\{x:z(x)>0\}$ containing $x_0$, inside the original coefficient interval. The first-order linear differential equation can have a longer existence interval than this inverse transformation allows.^[ref]

Negative values need a separately justified real branch. For integer $n$, the required powers are real and differentiable for $y<0$ too, but the inverse must retain the sign selected by the initial condition. For $n=3$, $z=y^{-2}>0$ represents either $y=+1/\sqrt{z}$ or $y=-1/\sqrt{z}$; both square roots are taken of positive $z$. For $n=2$, $z=1/y$ retains the sign directly. For a general noninteger exponent, use the positive branch unless a specific differentiable real negative branch has been established. Never evaluate a real fractional power of a negative value merely because the linear formula has become negative.

Division by $y^n$ excludes $y=0$. When $n>0$ and the original real power is defined at zero, $y\equiv 0$ is an additional solution of a differential equation and must be checked in the original equation. When $n<0$, the equation as written is undefined at zero. On a positive branch the right-hand side is locally Lipschitz in $y$, so the Picard-Lindelöf theorem gives local uniqueness. At zero, the same hypothesis needs a fresh check: it holds for the polynomial $n=2$ example, whereas for $0<n<1$ a nonzero power coefficient prevents local Lipschitz continuity there. If $z\downarrow 0$, the inverse sends $y\to \infty$ for $n>1$, but sends $y\downarrow 0$ for $0<n<1$. The latter endpoint can lead to finite-time extinction in a sublinear Bernoulli equation, where the original equation determines how to join to zero.

Under fixed-support regularity permitting differentiation under the integral sign, apply the product rule to $\int s_{\lambda }{W}_{\lambda }$. The first derivative term is $\int W_{\lambda }\nabla s_{\lambda }=\mathbb{E}[W_{\lambda }\nabla \log s_{\lambda }]$ and the second is $\mathbb{E}[\nabla W_{\lambda }]$. This proves the displayed identity. For an evidence lower bound, take $W_{\lambda }=\log p(t_{\lambda }(\epsilon ),x)-\log q_{\lambda }(t_{\lambda }(\epsilon ))$, holding noise $\epsilon$ fixed in the inner derivatives. Parameter-independent continuous noise contributes a reparameterization gradient; untransformed discrete choices contribute a score-function gradient estimator. A mixed base law cannot be called parameter independent if its discrete factors retain variational parameters.^[ref]

For bounded linear operators, define $C(\beta )=[A,e^{-\beta B}]$. The product rule gives $C^{\prime }+BC=[B,A]e^{-\beta B}$ and $C(0)=0$. Multiplying by the integrating factor $e^{\beta B}$ yieldsThe identity is useful when the two linear operators do not commute: it preserves their order rather than incorrectly differentiating an operator exponential as if all its factors were scalars. For unbounded linear operators, the products and differentiations need a common justified operator domain.

Let $X$ be a Conformal Killing vector field with conformal Killing factor $\lambda$, and let $U$ be the tangent vector of a geodesic with affine parameter $s$. The product rule, metric compatibility and geodesic equation giveEvery null geodesic therefore has the conserved quantity $g(X,U)$. On a timelike geodesic with proper time as parameter, the derivative is $-\lambda$ in metric signature $(-,+,+,+)$. With a nonaffine parameter satisfying ${\nabla }_{U}U=fU$, an extra term $fg(X,U)$ appears; the conserved expression for a null geodesic is tied to its affine parameter normalization.

With complex scalars and pointwise multiplication, the product rule gives a submultiplicative norm. For a Cauchy sequence, uniform limits $f$ of the functions and $g$ of their derivatives satisfy $f(t)-f(0)={\int }_0^{t}g(s)ds$, proving $f^{\prime }=g$ and completeness. Polynomials are dense in this norm: approximate $f^{\prime }$ uniformly by polynomials and integrate, fixing the initial value. This also identifies the character space of C1 on a compact interval.

For a continuously differentiable scalar field $\varphi$ and vector field $\mathbf{F}$, the curl satisfiesIndeed, with fixed Cartesian basis vectors ${\mathbf{e}}_i$, the Cartesian expression for the curl and the product rule giveThe first sum is the cross product of the gradient of $\varphi$ with $\mathbf{F}$; the second is $\varphi$ times the curl of $\mathbf{F}$. The formula is useful when applying the Stokes theorem to a weighted vector field.

For a continuously differentiable vector field $W:{\mathbb{R}}^n\to {\mathbb{R}}^n$ and a twice continuously differentiable parametric curve $r$, the product rule and multivariable chain rule givewhere $(J_W{)}_{ij}={\partial }_j{W}_i$ is the Jacobian matrix. Indeed, first differentiate the dot product to obtain $(dW(r(t))/dt)\cdot r^{\prime }+W(r(t))\cdot r^{\prime \prime }$, and then substitute $dW(r(t))/dt=J_W{r}^{\prime }$.

The first term describes variation in the sampled vector field; the second records the changing velocity vector of the parametric curve. In three dimensions one may take $W=\nabla \times F$ for a sufficiently smooth vector field $F$. The formula then involves first and second spatial partial derivatives of $F$ and the velocity and acceleration of the parametric curve. Omitting the second term would be valid only when its dot product vanishes, for example for a straight path with constant velocity.

For a continuous function $f:\mathbb{R}\to \mathbb{R}$, $g(t)=t^{2}f(t)$ is differentiable at zero with $g^{\prime }(0)=0$, because $[g(h)-g(0)]/h=hf(h)\to 0$. At any nonzero point $x$, $g$ is differentiable exactly when $f$ is: one direction uses the product rule, and the converse applies it to $f(t)=t^{-2}g(t)$ near $x$.

Apply the product rule to the scalar quadratic form and transpose its last scalar term: $x^{T}Adx=(dx{)}^T{A}^{T}x$. Only the symmetric part of $A$ contributes to its value or its derivative with respect to $x$. A varying matrix also contributes the independent term $x^T(dA)x$. No assumption that $A$ is symmetric is needed. Repeating this rule and the chain rule computes mixed derivatives when vector and matrix arguments depend on separate parameters.

For distinct real or complex numbers ${\lambda }_j$ and positive integers $m_j$, the vector space spanned by $t^k{e}^{{\lambda }_{j}t}$, $0\le k<m_j$, is preserved by differentiation. In each ordered basis $(e^{{\lambda }_{j}t},t{e}^{{\lambda }_{j}t},\dots )$, its matrix has ${\lambda }_j$ on the diagonal and $1,2,\dots ,m_j-1$ just above it. This follows from the product rule. Rescaling to $t^k{e}^{{\lambda }_{j}t}/k!$ makes each block a Jordan block. In particular, differentiation is invertible on this vector space exactly when every ${\lambda }_j\ne 0$: the determinant is ${\prod }_j{\lambda }_j^{m_j}$.

For $r=|x|>0$, the gradient satisfies ${\partial }_{i}r=x_i/r$. The product rule and Einstein notation then give the displayed divergence formula in dimension $n$. In dimension three, its derivative matrix is symmetric, so the curl vanishes. Zero divergence forces $f(r)=C{r}^{-n}$ on each positive radial interval. This formula does not assert regularity at the origin.

For continuously differentiable vector fields, the gradient of their dot product expands as displayed. The contraction of two Levi-Civita symbols gives $[F\times (\nabla \times G){]}_i=F_j{\partial }_i{G}_j-F_j{\partial }_j{G}_i$; adding the directional derivative cancels the second term. Interchanging the fields and using the product rule proves the identity. It does not require either field to be irrotational.

For a material line tangent $\mathbf{\ell }$, $D_t\mathbf{\ell }=L\mathbf{\ell }$ with velocity gradient $L_{ij}={\partial }_j{u}_i$. The product rule gives $D_t(\mathbf{\ell }{\mathbf{\ell }}^T)=L\mathbf{\ell }{\mathbf{\ell }}^T+\mathbf{\ell }{\mathbf{\ell }}^T{L}^T$. Consequently the upper-convected derivative vanishes for any transported sum of these outer products. Under the derivative-index-first convention $G=L^T$ the same formula is $D_{t}A=G^{T}A+AG$. The distinction prevents transposition errors in constitutive equations.

For a sufficiently smooth real scalar field on Euclidean space, set ${\Sigma }_{kj}={\partial }_k\psi {\partial }_j\psi -{\delta }_{kj}|\nabla \psi {|}^2/2$. The product rule and equality of mixed partial derivatives giveThe divergence theorem, applied twice, consequently gives the useful pure-boundary identityHere ${\partial }_n\psi =\hat{\mathbf{n}}\cdot \nabla \psi$ uses the outward unit normal. This is the Euclidean zero-potential counterpart of the scalar stress-energy divergence identity.

For a constant $a$ and a polynomial $P$, the product rule gives $D(e^{ax}v)=e^{ax}(D+a)v$. Iteration and linearity prove the displayed identity. In particular $(D-a{)}^m(e^{ax}v)=e^{ax}{v}^{(m)}$, which reduces a repeated-root constant-coefficient linear ordinary differential equation to repeated integration.

The exterior derivative is the unique graded derivation $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$ that agrees with the differential on functions and satisfies $d^2=0$.

On a smooth planar one-form $\alpha =Pdx+Qdy$, the product rule and antisymmetry of the wedge product of differential forms giveFor $\alpha =xdy$, the coefficient is $1$: there is a nonzero integral around the boundary of every positively oriented rectangle. For $\beta =d(xy)$, the coefficient is ${\partial }_{x}x-{\partial }_{y}y=0$. More generally, the identity $d^2=0$ makes every exact differential form a closed differential form. It expresses cancellation between mixed partial derivatives locally; it does not imply that every closed differential form has a global potential.

For a positive integer $m$, the negative matrix power is differentiable on the open general linear group. The Neumann series gives $(A+H{)}^{-1}=A^{-1}-A^{-1}H{A}^{-1}+O(\Vert H{\Vert }^2)$ locally around any invertible matrix. Applying the product rule to the $m$ factors of this inverse gives the displayed Fréchet derivative. Each summand chooses one inverse factor to differentiate while keeping the others in order.

The factor order matters. Only when $A$ and $H$ commute does the sum simplify to $-m{A}^{-(m+1)}H$. This is a useful example of why scalar differentiation rules for powers cannot be transferred to matrices without accounting for noncommutativity.

For $a>0$, $b\ne 0$, and a positive integer $\beta$, consider $x^2{y}^{\prime \prime }+x{y}^{\prime }+(-a^2+b{x}^{\beta })y=0$ near $x=0$. Its indicial roots are $a,-a$. The first Frobenius solution, normalized to leading coefficient one, is $y_{+}=x^a{\sum }_{k\ge 0}{c}_{k\beta }{x}^{k\beta }$, with $c_{k\beta }=-b{c}_{(k-1)\beta }/[k\beta (k\beta +2a)]$. If $N=2a$ is a positive integer, a second pure Frobenius solution with leading coefficient one would requireBefore $n=N$, induction makes $d_n=0$ for indices not divisible by $\beta$. If $\beta$ divides $N$, the coefficient $d_{N-\beta }$ is nonzero, and the exceptional equation $0=-b{d}_{N-\beta }$ is impossible. If $\beta$ does not divide $N$, the right-hand side is zero and $d_N$ is free. Thus, in the Frobenius case classification, a positive integer gap gives Case IIIa exactly when $\beta$ divides $2a$, and Case IIIb otherwise.

The coefficient of the necessary logarithm can also be calculated. Let $L=x^2{d}^2/d{x}^2+xd/dx-a^2+b{x}^{\beta }$. The product rule gives $L(y_{+}\log x)=2x{y}_{+}^{\prime }$. Therefore a second solution $x^{-a}\sum d_n{x}^n+K{y}_{+}\log x$, with $d_0=1$, hasFor $N=m\beta$ the preceding recurrence relation gives $d_{(m-1)\beta }=b^{m-1}/[{\beta }^{2m-2}((m-1)!{)}^2]$, soThis proves that the logarithm cannot disappear for a special nonzero value of $b$. For $a=0$, reduction of order instead produces the repeated-root Logarithmic Frobenius solution directly.