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Use the substitution $u=x^2$, so $du=2xdx$. The antiderivative isDifferentiation by the chain rule gives $4x{e}^{x^2}$, verifying the result.

Write $Y(t)=y(e^t)$, so $t=\log x$. The chain rule givesSubstitution into the Cauchy-Euler equation therefore gives $Y^{\prime \prime }+Y^{\prime }-2Y=0$. Its characteristic equation factors as $(r-1)(r+2)=0$, so $Y=C_1{e}^t+C_2{e}^{-2t}$. Returning to $x$ gives

The chain rule givesThe exponential function never vanishes, so all critical points areIn particular $f^{\prime }(0)=0$.

Differentiate the cumulative distribution function for $y>0$ using the chain rule and the standard normal density:Thus a probability density function isThe value assigned at zero is immaterial. The probability density function is unbounded near zero, but $F_Y(0)=0$ and continuity of the cumulative distribution function there show that there is no point mass. This is a gamma distribution with shape $1/2$ and rate $1$.

Apply the multivariable chain rule to a differentiable function $f(s,t)$ on the chosen local branch of the parabolic coordinates:Substituting part (ii) gives

The chain rule gives $\frac{d}{dx}\cos (\mu x)=-\mu \sin (\mu x)$. Dividing by the nonzero constant $-\mu$ gives the antiderivative

Write $Y(t)=y(e^t)$, so $x=e^t>0$. The chain rule givesHence $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$ and the Cauchy-Euler equation becomesIts characteristic equation is $r^2+3r+2=(r+1)(r+2)=0$. The distinct roots give $Y(t)=A{e}^{-t}+B{e}^{-2t}$. Returning to the original independent variable givesThe same form holds on any negative interval avoiding $0$, using $t=\log |x|$; the stated substitution covers the positive interval. At $x=0$ the differential equation is singular, so crossing it is a separate extension problem.

Reversing sign reverses the inequality. Since $X$ has a probability density function, it has no point masses, soDifferentiating the cumulative distribution function givesalmost everywhere: the minus sign outside $F_X$ cancels the minus sign from the chain rule. If point masses were allowed, the cumulative distribution function would instead require the left limit $1-F_X((-y{)}^{-})$.

Translate the inequality inside the cumulative distribution function:Differentiating with the chain rule gives the probability density functionalmost everywhere. Translation shifts the probability distribution without changing its shape or its total probability.

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 $u$.

The coordinate sum is $2\sin t$, while the sum of their cubes is $6\sin t-4{\sin }^{3}t$. Along the parametrized curve, the composite function is therefore $-3\sin t/2-{\sin }^{3}t/3$. Differentiate, or apply the multivariable chain rule, to obtain

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.

Use a column gradient. By the multivariable chain rule,The scalar coefficients multiply the constant vectors ${\mathbf{a}}_i$. In particular $f^{\prime }(t)=-t{e}^{-t^2/2}$, although explicit differentiation of $f$ is not required.

Put $\mathbf{z}=(x,y{)}^T$, $r^2={\mathbf{z}}^T\mathbf{z}$ and $f=\sqrt{1+r^2}$. The chain rule yieldsFor every nonzero vector $\mathbf{v}$, the Cauchy-Schwarz inequality givesThus the Hessian is a positive-definite matrix everywhere. The Hessian criterion for convexity implies that $f$ is strictly convex on ${\mathbb{R}}^2$. Its unique global minimum is at $\mathbf{z}=0$, where $f=1$.

Write $r=x-a$. The squared Euclidean norm is $r^{T}r$, whose gradient is $2r$. The chain rule for the exponential function therefore gives the column gradientThe row derivative, if that convention is used for $df/dx$, is its transpose.

Here a dense $100000$-by-$100000$ matrix has $10^{10}$ entries, about $80$ GB even with eight-byte storage per entry. Use limited-memory BFGS on the negative of the objective, or equivalently its positive-definite approximate-metric form of gradient ascent. It stores only a fixed small number $\ell$ of vector pairs, costing $O(\ell d)$ memory rather than $O(d^2)$.

Let $G(x)={\sum }_{i=1}^{100}{e}^{e^{-\Vert x-a_i{\Vert }^2}}$ and put $c_i(x)=e^{e^{-\Vert x-a_i{\Vert }^2}}{e}^{-\Vert x-a_i{\Vert }^2}$. Direct use of the chain rule givesIf $H_n$ denotes the positive-definite inverse-Hessian matrix approximation for $-G$, the iteration iswith a line search increasing $G$ and the usual safeguarded limited-memory updates. No exact Hessian matrix is required. With fixed memory and the usual bounded-metric and line search hypotheses, expect local linear convergence near a nondegenerate strict maximum; fixed-memory limited-memory BFGS does not generally inherit the superlinear theorem for full-memory BFGS.

Unlike the preceding objective, $G$ is not generally a concave function. Thus this is a method for finding a local maximum; comparison of different converged candidates or a global search is needed to certify a global maximum for arbitrary centres. The symbolic problem supplies no assumption that would justify an unconditional global-convergence claim.^[ref]

A pathline records the motion of one identified fluid element: keep its starting position $\mathbf{a}$ and starting time $t_0$ fixed, and let physical time $t$ advance. In a velocity field $\mathbf{u}(\mathbf{x},t)$ its Lagrangian trajectory $\mathbf{X}(t;t_0,\mathbf{a})$ solves the initial value problemA streamline instead freezes the time argument of the velocity field; a streakline fixes the observation time and varies the release time of many fluid elements. A pathline is useful for answering where a particular parcel goes and how long it takes, rather than merely which way the fluid is moving now. ^[ref]

For a concrete analytical model, take spatial coordinates $x,y$ in metres, time $t\ge 0$ in seconds, and the spatially uniform but time-dependent velocity fieldIts numerical components in these units are $(1,t)$. A fluid element starting at the origin at $t_0=0$ has the pathlineAt $t=1\mathrm{s}$ it is at $(1,0.5)\mathrm{m}$; at $t=2\mathrm{s}$ it is at $(2,2)\mathrm{m}$. The parabola records earlier, shallower velocity directions as well as the final direction. Differentiating gives $(\dot{x},\dot{y})=(U,at)$ and acceleration $(0,a)$, so this curved pathline needs no spatial variation of the velocity field. More generally the chain rule along a pathline gives $df(\mathbf{X}(t),t)/dt={\partial }_{t}f+\mathbf{u}\cdot \nabla f$, the material derivative of a transported scalar field.

A motivating tracer demonstration appears in S. J. Kline's film notes: water passes through a contracting channel, and hydrogen bubbles mark a small element at a wire at time $t_0$. The illustrated element is shown again $4.7\mathrm{s}$ later, with its accumulated pathline superposed. Following that same marked element across successive film frames records a history; the entire bubble pattern in one frame has a different meaning. ^[ref]

For a modern version of the readout, use sparse small markers in a transparent water channel, illuminate the viewed plane, and keep an optical camera fixed with a ruler in that plane. Record timestamps and track the same isolated marker between frames; successive positions estimate its pathline and $[\mathbf{X}(t+\Delta t)-\mathbf{X}(t)]/\Delta t$ estimates its velocity. A long exposure records a pathline segment but loses the timing within it, so its total length divided by exposure time gives only mean speed. These are proposed measurement steps based on Jean Hertzberg's teaching guidance, rather than additional measurements from Kline's film. Markers leaving the viewed plane, overlapping tracks and perspective can break particle identification; repeat with a thinner viewed slice and a lower marker concentration. ^[ref]

A phase modulation changes the local position of a periodic pattern. Its local wavenumber is shifted by derivatives of the phase. In a method of multiple scales, a correction profile depending on the translated coordinate $\xi =x+\varphi (Y)$ must be differentiated with ${\mathcal{D}}_Y={\partial }_Y+{\varphi }_Y{\partial }_{\xi }$. Omitting this chain rule can change the nonlinear coefficient of a reduced amplitude equation even when its linear coefficients are unchanged.

For a homogeneous polynomial $F$ of degree $d$, its first polar with respect to $[a:b:c]$ is the equation $a{F}_x+b{F}_y+c{F}_z=0$, of degree $d-1$ when this polynomial is nonzero. Every singular point of $V(F)$ lies on every such polar. Polars turn the Bézout's theorem intersection count into bounds on singularities and tangencies. The equation is covariant under a linear change of homogeneous coordinates, by the chain rule.

The change of reactor reactivity with thermal power along a specified family of states. In a quasistatic process, the chain rule weights each temperature coefficient of reactivity by that component's temperature response to power.