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The vector $m$ is constant, so $\nabla (m\cdot r)=m$. Also $\nabla (r^{-3})=-3r/r^5$, by the same chain rule. The product rule therefore yieldsThe three components follow by differentiating $(\alpha x+\beta y+\gamma z)(x^2+y^2+z^2{)}^{-3/2}$.

Use the product substitution for a first-order equation $v=xy$. The product rule gives $v^{\prime }=x{y}^{\prime }+y$, turning the equation into $v^{\prime }=e^v$. Consequently $(e^{-v}{)}^{\prime }=-1$, so $e^{-v}=C-x$.

At $x=1$ the initial value gives $v=1$, hence $C=1+e^{-1}$. Recovering $y=v/x$ givesThis is the maximal real graph interval containing the initial point: the logarithm requires $x<1+e^{-1}$, and the numerator has a nonzero limit at zero, so this particular solution cannot extend through zero. Differentiating $xy=-\ln (C-x)$ verifies $x{y}^{\prime }+y=1/(C-x)=e^{xy}$.

Put $s=b^a{t}^{a}x$ and write $y^{\prime }$ for differentiation with respect to $s$. Work on a real branch where the powers are defined and differentiable; $b>0$, $t>0$ suffices for arbitrary real $a$, and the parameter choice below has $b>0$. At fixed $x$, $\partial s/\partial t=as/t$. The product rule and chain rule therefore give

The product rule and chain rule give $y^{\prime }=\sin (x^2)+2x^2\cos (x^2)$. At zero the slope is zero and the curve passes through $(0,1)$. Consequently the tangent line is

The product rule gives the two partial derivativesThey both vanish at the origin. The exponential function is never zero, so any further stationary point must satisfy $x=0$ or $x^2-y^2=2$, and $y=0$ or $y^2-x^2=2$. Having both coordinates nonzero would require both opposite differences to be $2$, which is impossible. If $y=0$, the nonzero choices are $x=\pm \sqrt{2}$; if $x=0$, they are $y=\pm \sqrt{2}$. Hence the full list is

For $x\ne 0$, induction giveswhere each $P_n$ is a polynomial. This recurrence follows from the product rule and $d(1/x)/dx=-1/x^2$. Since Gaussian decay dominates every power,These limits also describe the derivatives of the extension at zero. If $f^{(n)}(0)=0$, its next difference quotient is $P_n(1/h)e^{-1/h^2}/h$, which again tends to zero. Induction proves existence and continuity of every derivative, all zero there. This is a flat function, covered more generally by polynomial multiples of a Gaussian flat function.

Apply the product rule and the chain rule to the exponential function:

For fixed amount of substance, take the internal energy $U=U(S,V)$. The first law of thermodynamics identifies $T=U_S{|}_V$ and $-P=U_V{|}_S$. Equality of mixed second derivatives givesFor the Helmholtz free energy, subtract the product differential $d(TS)$ from the first law of thermodynamics to obtain $dF=-SdT-PdV$. Its natural variables are $(T,V)$, so $F_T=-S$ and $F_V=-P$. Exactness therefore givesFor the Gibbs free energy, the product rule gives $dG=-SdT+VdP$. Thus $G_T=-S$, $G_P=V$, andFor the enthalpy, $dH=TdS+VdP$, giving $H_S=T$, $H_P=V$ andThese are the four Maxwell relations, each following from exactness in the appropriate pair of natural variables.

First take $F_y=x{e}^{\sin (xy)}\cos (xy)$, then differentiate with respect to $x$. The product rule and chain rule giveThe function is smooth, so the two mixed partial derivatives agree.

Differentiate the polar form of a complex number using the product rule and chain rule:A second differentiation givesThe two contributions to $2\dot{r}\dot{\theta }$ arise separately from differentiating $r\dot{\theta }$ and from differentiating the moving exponential factor.

For the first-order linear ordinary differential equation, seek an integrating factor $M(x)$ that makes its left side a product rule derivative. Multiplying by $M$ gives $M{y}^{\prime }+MPy$, and comparison with $(My{)}^{\prime }=M{y}^{\prime }+M^{\prime }y$ requires $M^{\prime }=PM$. On an interval where $P$ is continuous chooseThis factor is never zero. The transformed equation is $(My{)}^{\prime }=MQ$, so integration from $x_0$ to $x$ yieldsHere $C=y(x_0)$ for an initial value problem. The indefinite integral version is $y=e^{-\int Pdx}[C+\int Q{e}^{\int Pdx}dx]$, with any constant choices inside the integrating factor absorbed into $C$. Continuity of $P,Q$ is a convenient sufficient hypothesis for a classical differentiable solution of a differential equation.

Since $1+x^2>0$ for all real $x$, divide the equation by that factor. Its linear coefficient can be integrated by polynomial division:Thus a positive integrating factor is $M=e^{3x^2/2}(1+x^2{)}^{-3/2}$. Multiplying the divided equation gives the product rule formThe integration by substitution $u=1+x^2$ givesThe initial condition $y(0)=1$ fixes $C-2=1$. The required solution is thereforeNo real singularity was introduced by the division or the integrating factor, so the solution extends over the entire real line.

For $x\ne 0$, divide by $x^2$ to obtain $y^{\prime }+y/x=1/x^2$, a first-order linear ordinary differential equation. An integrating factor is $x$ on either interval not crossing zero, since the product rule givesIntegration and division by $x$ give the explicit general solutionThe original equation would read $0=1$ at zero, so no solution can include that point.

The product rule gives $y^{\prime }=e^x(\cos x-\sin x)$. The exponential function never vanishes, so a stationary point requires $\tan x=1$. In the given interval this means $x=\pi /4$. Substitution givesThe derivative changes from positive to negative there, making it a strict local maximum.

Write $c=\cos \varphi$ and $s=\sin \varphi$. The needed coefficient derivatives are $c_x=s^2/r$, $s_x=-sc/r$, $c_y=-sc/r$ and $s_y=c^2/r$. Apply the product rule to both terms of $f_x=c{g}_r-(s/r)g_{\varphi }$ and use the chain rule for the derivatives of $g$. Collecting terms givesSimilarly, differentiating $f_y=s{g}_r+(c/r)g_{\varphi }$ givesOn addition, both the mixed partial derivative terms and the first angular derivative terms cancel. Using $s^2+c^2=1$, the Laplacian in polar coordinates is

Under $y=xu$, the product rule gives $y^{\prime }=u+x{u}^{\prime }$ and $y^{\prime \prime }=2u^{\prime }+x{u}^{\prime \prime }$. Substitute these into the original equation:Since $x>0$, this is precisely $x{u}^{\prime \prime }+u^{\prime }=0$, or $(x{u}^{\prime }{)}^{\prime }=0$. Integrating twice gives $x{u}^{\prime }=C_1$ and $u=C_1\ln x+C_2$. ThereforeThe two independent solutions of a differential equation supply the general solution of this second-order linear ordinary differential equation.

Again write the field as $\mathbf{b}(\mathbf{a}\cdot \mathbf{r})-\mathbf{r}(\mathbf{a}\cdot \mathbf{b})$. The second term has zero curl. Applying the product rule to the first gives $\nabla (\mathbf{a}\cdot \mathbf{r})\times \mathbf{b}=\mathbf{a}\times \mathbf{b}$. Thus

For $x\ne 0$, multiply by $x$ to obtain the first-order linear ordinary differential equation $y^{\prime }+xy=5x{e}^{-x^2}$. Its integrating factor is $\mu =e^{x^2/2}$. The product rule then givesIntegration yields $y{e}^{x^2/2}=-5e^{-x^2/2}+C$, soThe constant of integration is arbitrary on each connected interval of the original equation's domain. The formula extends smoothly through zero, but the printed equation itself contains $1/x$ and is not defined there.

The characteristic equation has a repeated root $-1$, suggesting the substitution $y=e^{-x}v$. By the product rule,so $y^{\prime \prime }+2y^{\prime }+y=e^{-x}{v}^{\prime \prime }$. The inhomogeneous linear differential equation reduces to $v^{\prime \prime }=2x$. Two integrations giveThereforeThe two arbitrary constants account for the repeated-root homogeneous solution.

Multiplication by the integrating factor and the product rule turn the left side into a single derivative:Using integration by substitution $u=\cos x$ gives $y\sin x=-{\cos }^{3}x/3+C$. Therefore the general solution isThere is one arbitrary constant of integration, as expected for a first-order ordinary differential equation.