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If $Lf=\lambda f$ and $f_t+Af=0$, differentiation and the Lax equation give $(L-\lambda )f_t=-A(L-\lambda )f$, so ${\lambda }_t=0$ after projection onto the normalized eigenfunction for a nondegenerate eigenvalue.

For $\Psi =(f,f_x{)}^T$,The compatibility ${\Psi }_{xt}={\Psi }_{tx}$ is precisely ${\partial }_t{M}_L-{\partial }_x{M}_A-[M_A,M_L]=0$. If all coefficients are $x$-independent, this becomes ${\dot{M}}_L=[M_A,M_L]$, and cyclicity of trace gives $d\mathrm{Tr}(M_L^k)/dt=0$.

Let $I={\int }_0^{\infty }{e}^{-x^2}dx$. Positivity allows the squared integral to be written as a first-quadrant double integral. In polar coordinates the Jacobian determinant is $r$, soUsing integration by parts, with the boundary term $x{e}^{-x^2}$ vanishing at both endpoints,The substitution $u=x^2$ also gives the two antiderivativesThese values also follow from the Gaussian moment recurrence and can be checked by differentiation.

Let $a=f(b,c)$ and work where both $f_b$ and $f_c$ are nonzero, so the required inverse partial derivatives exist. At fixed $a$, the total differential gives $0=f_{b}db+f_{c}dc$, hence $(\partial b/\partial c{)}_a=-f_c/f_b$. At fixed $b$, $(\partial c/\partial a{)}_b=1/f_c$. Their product with $f_b$ is $-1$: this is the cyclic partial derivative identity.

For explicit verification in this example, put $P=b^4-6b^2{c}^2+c^4$ and $H=(b^2+c^2{)}^3$. Direct differentiation givesThereforeThe identity is local to regular branches: for these explicit formulas require $bcP\ne 0$. At singular points one cannot justify cancellation of nonexistent inverse derivatives merely by writing a formal product.

Direct differentiation of $v=(x^2-1{)}^l$ gives $(x^2-1)v^{\prime }=2lxv$; the identity also holds for $l=0$. Differentiate it $l+1$ times using the Leibniz rule. The quadratic factor has only its first two derivatives nonzero, soWriting $y=v^{(l)}$ and collecting terms gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$. Multiplication by $-1$ gives the Legendre differential equation:

At fixed $t$, the argument is linear in $x$, with $\partial s/\partial x=b^a{t}^a$. Each differentiation therefore contributes this same constant factor, while the outside factor $t^a$ remains unchanged. ThusHere $y^{(3)}$ denotes the third derivative of the single-variable profile, not a partial derivative of $u$.

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

Differentiate the Fourier series term by term and collect cosine and sine terms. Substitution into the equation givesFor a smooth periodic solution, differentiation of the series is justified. Alternatively, multiply the differential equation by each trigonometric mode and integrate by parts; periodicity cancels the boundary terms and gives the same coefficient constraints without first differentiating an infinite sum.

The absolute value in the original PDF reflects every negative sine arch upward. The graph is nonnegative, with zeros at $x=k\pi$ for $k=0,1,2,3,4$, and it lies below the decaying envelope $e^{-x}$. On the $k$th arch, differentiation gives a factor proportional to $\cos x-\sin x$. Thus its unique local maximum isSuccessive maximum heights have ratio $e^{-\pi }$. At each interior zero the one-sided slopes have opposite signs, producing a corner. The smaller panels below magnify the later arches so that all four remain visible.

The general solution of this first-order linear ordinary differential equation is $y=C{e}^{3x}$. Its initial condition gives $C=3$, soDifferentiation verifies $y^{\prime }=3y$ and substitution at zero gives the required value.

Normalize the Rayleigh distribution by substituting $s=x^2/(2{\sigma }^2)$:For its sketch, the density starts at zero, is positive for $x>0$, and tends to zero at infinity. Differentiation givesIt therefore has a single maximum at $x=\sigma$, with height $e^{-1/2}/\sigma$.

The expected value isIntegrating by parts with $xdx{e}^{-x^2/(2{\sigma }^2)}=-{\sigma }^{2}d(e^{-x^2/(2{\sigma }^2)})$ makes the boundary term vanish and leaves ${\int }_0^{\infty }{e}^{-x^2/(2{\sigma }^2)}dx$. The Gaussian integral gives

The sum is finite, so differentiation passes through it without a convergence issue. Each lower limit $n$ is constant with respect to $x$, and the fundamental theorem of calculus gives the same upper-end integrand. Hence the derivative isThe binomial theorem at $1+1$ gives the sum $2^N$. Therefore

The first field has vanishing curl and is the gradient of a monomial:For the second field, direct component differentiation givesIt is not identically zero. Thus ${\mathbf{F}}_1$ is conservative and ${\mathbf{F}}_2$ is not on ${\mathbb{R}}^3$. The conditional path-integral requests therefore apply to ${\mathbf{F}}_1$.

Substitution first makes the square-root branch transparent:On the given interval this is exactly $\sin x$, so direct differentiation givesin agreement with the chain-rule calculation. Outside this interval one must retain the absolute value; there are cusps at integer multiples of $\pi$.

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.

Write $\mathbf{a}=(a_x,a_y,a_z)$. The cross product gives ${\mathbf{v}}_1=(a_{y}z-a_{z}y,a_{z}x-a_{x}z,a_{x}y-a_{y}x)$, whose three diagonal partial derivatives vanish. For the second vector field, direct differentiation gives ${\partial }_x(2xy)=2y$, ${\partial }_y(x^2+3y^2+z^2)=6y$, and ${\partial }_z(2yz)=2y$. Thus

Separate variables near the nonzero initial condition: $y^{-2}dy=dx$. Integration gives $-1/y=x+C$, and $y(1)=1$ fixes $C=-2$. ThereforeThis is the maximal connected interval containing the initial point. Direct differentiation gives $y^{\prime }=1/(2-x{)}^2=y^2$; the pole at $x=2$ prevents continuation as a finite classical solution of a differential equation through that point.

The positive-domain graph starts with limiting value $f(0^{+})=1$ and limiting slope $-1$, then decreases towards the horizontal asymptote $f=0$, with $f(q)\sim 1/(2q)$ for large $q$. In fact direct differentiation givesThe first inequality follows from $e^s>1+s$ for $s>0$. Thus the graph is strictly decreasing and convex upward, with $0<f(q)<1$ and no stationary points. The integral representation $f(q)={\int }_0^1(1-e^{-s/q})ds$ also shows its positivity.

These properties belong to the reusable exponential spherical mass factor.

Put $D={\sigma }^2>0$ and $\xi =x-x_0$. Logarithmic differentiation of the positive Gaussian function givesDifferentiating once more in space givesMultiplying by $D$ provesThe assumption $\sigma \ne 0$ ensures both the normalization and exponential denominator are defined. This is the heat kernel, with diffusion coefficient $D$ and centre $x_0$.

On the negative half-line, differentiation gives $d\ln (-u)/du=1/u$. The integration path from $-1$ to $x<0$ stays away from the singularity at zero. ThereforeThis choice satisfies $F(-1)=0$.

Let $H(u)={\sin }^{2}u+\ln u$. The chain rule gives $H^{\prime }(u)=\sin 2u+1/u$, so the two given summands combine into its logarithmic derivative, $f(u)=H^{\prime }(u)/H(u)$. For $u>1$, $H(u)>0$, so the real logarithm is defined throughout the integration path. Since $H(\pi )=\ln \pi$,This satisfies $F(\pi )=0$, and differentiation recovers both terms of the integrand.