The generalized Leibniz rule follows by mathematical induction. The case $n=0$ is the identity $uv=uv$. Suppose the formula holds for $n$ and differentiate every term using the product rule:For an interior index $r$, the coefficient of $u^{(r)}{v}^{(n+1-r)}$ becomes $\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)+\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ by Pascal's identity. The two endpoint coefficients are one. This provesOnly finitely many derivatives are involved; it suffices that both functions are $n$ times differentiable.

For $u=x(x+1)$ and $v=e^{2x}$, the derivatives of $u$ beyond order two vanish. The three surviving terms of the generalized Leibniz rule giveThe simplified expression works for every integer $n\ge 0$, including the two cases where one or more of the three terms are absent.

For $u=x^2$ and $v=\sin x$, use $d^m(\sin x)/d{x}^m=\sin (x+m\pi /2)$. Again only three terms survive. With $T=x+n\pi /2$, they areSince $\sin (T-\pi /2)=-\cos T$ and $\sin (T-\pi )=-\sin T$, this becomesThe factors $n$ and $n(n-1)$ remove the nonexistent terms when $n=0$ or $1$.

If $Pdx+Qdy=dF$ for a twice continuously differentiable potential $F$, then $P=F_x$ and $Q=F_y$. Equality of mixed partial derivatives gives the necessary condition for an exact differential:This condition alone does not ensure global exactness on a domain with holes.

For the Helmholtz free energy $A=U-TS$, use the stated identity for internal energy and the product rule:Thus $A_T{|}_V=-S$ and $A_V{|}_T=-p$. Equality of mixed partial derivatives yields the Maxwell relationThe temperature and volume are the independent variables here, with composition and amount of gas fixed.

Next express $S=S(T,V)$ and substitute its total differential into $dU=TdS-pdV$:Matching the coefficient of $dV$ givesThis is the thermodynamic internal pressure identity: an equation of state determines how internal energy changes under an isothermal change of volume.

For the Dieterici equation, set $c=an/(VR)$ while differentiating at fixed $V$. ThenSubstitute into the thermodynamic internal pressure identity. The term proportional to $T$ cancels $p$, leavingThe physical gas domain has $T>0$ and $V>nb$.

For the ideal gas instead, $p=nRT/V$ gives $T(\partial p/\partial T{)}_V-p=0$. Therefore $U$ is independent of volume at fixed amount of gas. Integrating the remaining condition $(\partial U/\partial T{)}_V=C_V$, with constant heat capacity at constant volume, giveswhere $U_0$ is independent of both $T$ and $V$ and fixes the reference for internal energy.

For an $N+1$ times continuously differentiable function, Taylor's theorem giveswhere $\xi$ lies between $x_0$ and $x$. An infinite Taylor series represents the function when its remainder tends to zero; this holds locally for an analytic function, rather than for every smooth function automatically.

Write $r=|\mathbf{r}|$ and $q=\mathbf{p}\cdot \mathbf{r}$. Away from the origin, $\nabla q=\mathbf{p}$ and $\nabla r^{-m}=-m\mathbf{r}/r^{m+2}$. The product rule givesConsequently the divergence of $\mathbf{E}=3q\mathbf{r}{r}^{-5}-\mathbf{p}{r}^{-3}$ is zero. For any closed surface bounding a region that excludes the origin, the field is smooth throughout that region, and the divergence theorem provesFor the infinite cylinder, use $\mathbf{r}=(a\cos \varphi ,a\sin \varphi ,z)$ and outward unit normal $(\cos \varphi ,\sin \varphi ,0)$. With $\mathbf{p}=(0,0,1)$, $q=z$, $dS=ad\varphi dz$, and $\mathbf{E}\cdot \mathbf{n}=3az/(a^2+z^2{)}^{5/2}$. Hence the direct surface flux isThe integrand is odd in $z$ and absolutely integrable. This calculation treats the unbounded cylindrical side directly, without applying the bounded-region divergence theorem across the field's singularity.