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A sufficient differentiation under the integral sign theorem is the following local domination criterion. Suppose $u(t,\cdot )$ is measurable for each $t$, $u(t_0,\cdot )$ is integrable at one point of a neighborhood $J$ of $t_0$, and for almost every $x$ the function $u(\cdot ,x)$ is differentiable on $J$. Assume an integrable function $g_J$ satisfies $|{\partial }_{t}u(t,x)|\le g_J(x)$ for $t\in J$ and almost every $x$. Then $I$ is differentiable on the interior of $J$ andThe mean value theorem bounds the difference quotients by $g_J$, and the dominated convergence theorem passes their pointwise limit through the integral. Applying this criterion locally throughout $(a,b)$ proves differentiability there. One may impose the stronger condition that every $u(t,\cdot )$ is integrable from the outset.

Differentiability at $x_0$ means that a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ exists withThe map $A=Df(x_0)$ is the total derivative. A linear map between these finite-dimensional spaces is bounded, so $\Vert f(x_0+h)-f(x_0)\Vert \le \Vert A\Vert \Vert h\Vert +o(\Vert h\Vert )\to 0$. Thus differentiability implies continuity.

For the displayed two-variable function,Writing $r=\sqrt{x^2+y^2}$ gives $|f(x,y)|/r\le r\to 0$. Therefore it is differentiable at the origin, with derivative zero.

Finally, put $h=x-x_0$, write $f(x_0+h)=Ah+r(h)$ with $r(h)=o(\Vert h\Vert )$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,By the Cauchy-Schwarz inequality the last two terms are $o(\Vert h\Vert )$: $A$ is bounded and $g(x_0)+s(h)$ stays bounded. HenceOnly continuity of $g$ at $x_0$ is needed, because the vanishing of $f(x_0)$ removes the term which would otherwise require differentiating $g$.

The maximum-likelihood estimator is a maximizerWriting ${\ell }_n=\log L_n$ and the score $S_n={\nabla }_{\theta }{\ell }_n$, the expected Fisher information matrix iswhere the equality uses the usual regularity conditions. For independent identically distributed observations, ${\mathcal{I}}_n=n{\mathcal{I}}_1$.

Under the standard identifiability, interior-parameter, differentiability and nonsingular-information assumptions, the large-sample distribution isor approximately $\hat{\theta }\sim N(\theta ,{\mathcal{I}}_n(\theta {)}^{-1})$. Boundary parameters need not obey this regular approximation.

Assume the regular Sturm–Liouville setting: real coefficients with $p>0$, $w>0$ and sufficient continuity and differentiability for the integrations below. Apply a multiplier to $J[y]-\lambda ({\int }_a^{b}w{y}^2-1)$. Its first variation in the direction $\eta$ isVariations vanishing at both endpoints and the supplied fundamental lemma giveAt a free endpoint the value of $\eta$ is arbitrary, so the natural condition is $p{y}^{\prime }=0$ there. At a fixed endpoint $\eta =0$, so no such derivative condition is imposed; the prescribed endpoint value is imposed instead. This distinction also allows one endpoint of each type.

For eigenfunctions $u,v$ of distinct eigenvalues $\lambda ,\mu$ under the same homogeneous self-adjoint endpoint conditions, multiply their equations by $v,u$ and subtract. Integration givesThe boundary expression vanishes for both natural conditions and homogeneous fixed-value conditions. Thus $\boxed{{\int }_a^{b}wuv=0}$ for different eigenvalues.

In the specified fixed-endpoint case the eigenproblem is $-y^{\prime \prime }-y=\lambda y$, $y(0)=y(1)=0$. There are no nonzero solutions when $\lambda +1\le 0$; the elementary exponential or affine solutions and the two endpoint conditions prove this. For $\lambda +1>0$, the first condition leaves a sine and the second requires frequency $n\pi$. The normalized stationary functions and eigenvalues areDistinct sine modes are orthogonal by the product-to-sum formula, or by the preceding Green second identity. Although they need not satisfy the natural derivative conditions, their fixed zero values also annihilate the boundary expression, so orthogonality is expected.

The Wirtinger inequality ${\int }_0^1{y}^{\prime 2}\ge {\pi }^2{\int }_0^1{y}^2$ gives $J[y]\ge {\pi }^2-1$. For completeness it follows by writing $y=v\sin \pi x$ for a smooth function supported away from the endpoints and integrating the cross term: $\int (y^{\prime 2}-{\pi }^2{y}^2)=\int {\sin }^2(\pi x)v^{\prime 2}\ge 0$. Approximation extends it to the Dirichlet Sobolev class. Equality is attained by the first sine mode; the eigenproblem or equality condition gives only its scalar multiples. Hence the two normalized minimizers are $\boxed{y=\pm \sqrt{2}\sin \pi x}$ and the minimum is $\boxed{{\pi }^2-1={\lambda }_1}$.

There is a wording distinction in the source: the two sign choices of the minimum are not orthogonal to each other, since their inner product is $-1$. Orthogonality applies to the stationary modes with different eigenvalues, not to both signs of the same minimizer.

The $i$th partial derivative is the limit ${\partial }_{i}f(x_0)={\lim }_{t\to 0}[f(x_0+t{e}_i)-f(x_0)]/t$. For a nonzero vector $v$, use the directional convention ${\partial }_{v}f(x_0)={\lim }_{t\to 0}[f(x_0+tv)-f(x_0)]/t$, without normalizing $v$. Differentiability means $f(x_0+h)=f(x_0)+Lh+o(\Vert h\Vert )$ for a linear map $L=df(x_0)$.

For the given example take $v=(v_1,v_2)$. If $v_1\ne 0$, then $f(tv)=t^2{v}_2^3/v_1$, so its directional difference quotient tends to zero. If $v_1=0$, the function vanishes along that line. Thus $\boxed{{\partial }_{v}f(0)=0\text{ for every }v\ne 0}$. However along $(x,y)=(t^3,t)$ with $t\ne 0$, one has $f(x,y)=1$, while $f(0,0)=0$. Hence the function is neither continuous nor differentiable at zero. Existence of all directional derivatives, even with a linear dependence on the direction, does not by itself imply total differentiability.

Define the scalar function $H(s)=⟨f(x+s(y-x)),z⟩$ on $[0,1]$. It is continuous there and differentiable inside, with derivative $⟨df(x+s(y-x))(y-x),z⟩$ by the chain rule. The one-dimensional mean value theorem gives $H(1)-H(0)=H^{\prime }(\theta )$ for some $0<\theta <1$, proving the displayed vector-paired version. No claim of a single intermediate point giving a vector equality without pairing is required.

For the final assertion, use the scalar mean value theorem directly on $s\mapsto g(sh)$: it is continuous at $s=0$ and differentiable for $s>0$. Thusfor some $0<{\theta }_h<1$. The assumed limits of all partial derivatives imply $\Vert \nabla g(x)\Vert \to 0$ as $x\to 0$, $x\ne 0$. Consequently $|g(h)-g(0)|/\Vert h\Vert \le \Vert \nabla g({\theta }_{h}h)\Vert \to 0$. Hence $\boxed{dg(0)=0}$ and $g$ is differentiable there. This argument uses only differentiability on the interior of the line segment, so it does not assume the conclusion at its endpoint.

The $k$th derivative of the integrand is $(\cos x{)}^k\cos (t\cos x+k\pi /2)$, whose absolute value is at most one. The integration interval has finite measure. Repeated application of the differentiation theorem therefore givesDominated convergence also gives continuity of every derivative, proving infinite differentiability.

A direct derivative bound by $1/x$ would not be integrable, so the earlier theorem cannot be applied that way. Instead, for $t>0$ make the legitimate substitution $s=tx$ in the absolutely convergent integral:The tail integral has continuous integrand away from zero and converges absolutely. The fundamental theorem of calculus therefore gives derivative $-\cos t/t^2$ with respect to its lower endpoint. The product rule yieldsThis establishes differentiability carefully without an invalid domination argument.

Differentiability at $x_0$ means that a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ exists such that $f(x_0+h)=f(x_0)+Lh+o(\Vert h\Vert )$. Since finite-dimensional linear maps are bounded, the right side tends to $f(x_0)$ as $h\to 0$, proving continuity.

For the specified scalar function, $0\le f(x,y)\le x^2\le x^2+y^2$. Dividing by $\sqrt{x^2+y^2}$ gives a bound tending to zero at the origin. Consequently $\boxed{df(0,0)=0}$ and the function is differentiable there.

For the final product assertion, write $f(x_0+h)=Lh+r(h)$, where $r(h)=o(\Vert h\Vert )$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,Cauchy–Schwarz and boundedness of $g$ near $x_0$ justify the last estimate. Thus $\boxed{dF(x_0)h=⟨df(x_0)h,g(x_0)⟩}$; differentiability of $g$ is unnecessary because $f(x_0)=0$.

The likelihood is $L(\theta ;x)={\prod }_{i=1}^{n}f(x_i;\theta )$, and a maximum likelihood estimator is any maximizer $\hat{\theta }\in \arg {\max }_{\theta }L(\theta ;X)$. Write $\ell =\log L$ and score $U={\nabla }_{\theta }\ell$. Fisher information for the sample iswhere the equality uses the usual differentiability and interchange assumptions. For independent identically distributed observations $I_n=n{I}_1$. Under standard interior, identifiability and regularity assumptions,Without those assumptions the large-sample normal assertion need not hold, for example at a boundary parameter.

Assume sufficient decay and differentiability to justify integration by parts and exchange of integrals; Schwartz functions are one convenient class. With the stated Fourier transform convention, integrate first in $x$:The boundary contribution vanishes under the decay assumption. The Fourier transform of a derivative consequently turns constant-coefficient spatial differential operators into multipliers. Repeating it gives $\mathcal{F}[u_{xx}]=-k_x^2\hat{u}$, so the heat equation becomes an independent temporal ordinary differential equation at each wavevector.

A Peano zero through order $r$ means that the function admits the zero Taylor polynomial with remainder $o(|t{|}^r)$ at $x_0$. This does not assert the existence of ordinary iterated derivatives on a neighborhood. For example, multiplying $t^{r+1}$ by a bounded continuous function with dense points of nondifferentiability preserves this estimate while destroying neighborhood differentiability.

A risk function can decay rapidly for each fixed regression function while its supremum over a function class decays more slowly or fails to decay. Bounds depending on each function's local differentiability remainder need not be uniform. In fixed-design nonparametric regression, a class with no common smoothness bound permits smooth bump functions whose supports shrink between design points. Such sequences obstruct uniform estimation without contradicting fixed-function mean squared error bounds.

Rolle's theorem is the equal-endpoint case of the mean value theorem. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, there exists $c\in (a,b)$ with $f^{\prime }(c)=0$. Geometrically, a tangent line somewhere in the interior is horizontal, like the secant line joining the endpoints. ^[ref]

For the polynomial $f(x)=x(x-2)$ on $[0,2]$, both endpoint values are zero and the derivative is $f^{\prime }(x)=2x-2$. Thus $c=1$, where $f(1)=-1$, and the horizontal tangent line is $y=-1$. The horizontal secant line is $y=0$: equal endpoint values do not force the interior tangent line to pass through those endpoints.

The absolute value function $f(x)=|x|$ on $[-1,1]$ exposes the importance of differentiability. It is continuous and $f(-1)=f(1)=1$, so the secant line has slope zero. However,and the derivative does not exist at zero because the one-sided derivatives disagree. Every existing derivative in the interior is nonzero. The corner fails a hypothesis of both Rolle's theorem and the mean value theorem, so neither conclusion is guaranteed.

For a short proof, the extreme value theorem supplies a greatest and a least value on $[a,b]$. If $f$ is a constant function, every interior point works. Otherwise at least one of these values differs from the common endpoint value, so it is attained at an interior extremum. At an interior local maximum or local minimum, differentiability forces the derivative to be zero: the difference quotients from the two sides have opposite weak signs and the same limit.

This turns repeated values into information about derivatives. Two distinct roots of a polynomial force a root of its derivative between them; iterating this idea gives the Rolle root count with multiplicities. More generally, a differentiable function with nonzero derivative throughout a real interval cannot repeat a value there, hence is injective. The beauty is that knowing only two equal values guarantees an exact local feature somewhere between them.

Smoothness describes the degree of differentiability of a function. The class $C^k$ has continuous derivatives through order $k$; a smooth function is of class $C^{\infty }$.

Let $n\ge 1$. Suppose $f$ has $n-1$ continuous derivatives on the closed interval joining distinct real points $a,x$, and its $n$th derivative exists in the interior. Then some interior point $\xi$ satisfiesEndpoint derivatives in the polynomial are one-sided restrictions of the continuous derivatives. When $n=1$, the assumptions mean endpoint continuity and interior differentiability, and the result is the mean value theorem.^[ref]

For a proof, subtract the degree-$(n-1)$ Taylor polynomial and a multiple $C(t-a{)}^n$ chosen so that the resulting function also vanishes at $x$. It and its first $n-1$ derivatives vanish at $a$. Repeated applications of Rolle theorem give a point where its $n$th derivative vanishes, yielding $C=f^{(n)}(\xi )/n!$. If $|f^{(n)}|\le M$, the Lagrange remainder is bounded by $M|x-a{|}^n/n!$ without knowing $\xi$.

The total differential is the linear map that combines the first-order responses to simultaneous changes in all coordinates. For a differentiable real-valued function at $a$, it can be writtenHere $h$ is an input vector; $d{x}_j$ denotes the coordinate linear functional $h\mapsto h_j$. The coordinate formula and the Fréchet derivative describe the same linear map. The defining approximation isThe little o notation says the remaining approximation error becomes small relative to the input norm, across every direction of approach.^[ref]

For $f=x^2+xy+2y^2$ at $a=(1,1)$, $d{f}_a(h,k)=3h+5k$. A simultaneous change $(h,k)=(0.02,-0.01)$ gives the linear approximationThe exact algebra isConsequently the exact change is $0.0104$, and the approximation error is $0.0004$. If both input changes are halved, the linear prediction becomes $0.005$ and the error becomes $0.0001$: the first-order response halves, while the quadratic error falls by a factor of four. The right panel under directional derivative displays this comparison as the displacement is scaled.

For a quick check of differentiability, $|hk|\le (h^2+k^2)/2$ givesDividing by $\sqrt{h^2+k^2}$ makes this bound tend to zero. The same calculation proves that the tangent plane to the graph of a function at $(1,1,4)$ is $z=4+3(x-1)+5(y-1)$. It approximates the entire local surface, rather than only two coordinate slices.

The total differential is useful for estimating the combined effect of small input changes without reevaluating a complicated function. For example, if $|h|\le 0.02$ and $|k|\le 0.01$, then $|d{f}_a(h,k)|\le 0.11$. The exact remainder bound from the displayed polynomial is at most $0.0008$, so this example has $|\Delta f|\le 0.1108$ throughout that input rectangle. These are deterministic bounds; a variance calculation additionally needs a joint probability distribution and its covariances. A first-order estimate is also local: large changes require control of the remainder, not merely knowledge of the partial derivatives at one point.

For an integrable periodic function of period $P>0$, its integral over an interval of length $P$ is independent of the starting point. Write $a=mP+r$ with $m$ an integer and $0\le r<P$. By a change of variables and the defining identity of a periodic function,No differentiability is required. This also allows the Fourier coefficients of a periodic function to be computed on any interval of one full period.

A vector-valued function is a function $F:X\to V$ taking values in a vector space. For example, $F(t)=(\cos t,\sin t)$ describes the unit circle in ${\mathbb{R}}^2$: one real number controls a two-component vector.

In a real or complex finite-dimensional vector space equipped with a norm, choosing a basis writes $F={\sum }_j{f}_j{e}_j$ in scalar-valued functions. Continuity and differentiability can then be checked in these coordinates, and $F^{\prime }={\sum }_j{f}_j^{\prime }{e}_j$ when the derivatives exist. Finitely many coordinates matter: coordinatewise convergence in an infinite-dimensional vector space need not imply norm convergence.

On $[0,1]$, the polynomials $p_n(x)=x^n/n$ converge in supremum norm to the zero function, since $\Vert p_n{\Vert }_{\infty }=1/n$. But $p_n^{\prime }(1)=1$ for every $n$, whereas the zero function has left derivative zero at that endpoint. Thus uniform convergence of function values does not imply uniform convergence of derivatives. This does not contradict uniform convergence of Bernstein polynomial derivatives, which has a particular approximation construction and additional differentiability hypotheses.