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A continuous bilinear map obeys $\Vert f(h,k)\Vert \le C\Vert h\Vert \Vert k\Vert$. Expanding $f(a+h,b+k)$ leaves the displayed linear term and remainder $f(h,k)$. For product norm $\rho =(\Vert h{\Vert }^2+\Vert k{\Vert }^2{)}^{1/2}$, the remainder has norm at most $C{\rho }^2/2$, hence is $o(\rho )$. This proves the Fréchet derivative formula.

For real rectangular matrices, expanding $(A+H{)}^t(A+H)$ leaves the quadratic remainder $H^{t}H$, whose matrix norm is $O(\Vert H{\Vert }^2)$. Thus the displayed Fréchet derivative is a linear map in the increment $H$. At an orthogonal matrix, its kernel consists of increments $H=A\Omega$ with $\Omega$ skew-symmetric. This is the tangent-space condition obtained by differentiating $A^{t}A=I$.

Differentiability is the existence of a derivative at a point. For maps between finite-dimensional normed vector spaces, the existence of a Fréchet derivative requires a single linear map approximating the increment with error $o(\Vert h\Vert )$. Merely having partial derivatives need not provide this approximation; continuous partial derivatives on a neighbourhood do suffice.

For $p>n$, the Hölder continuous function representative $u^{\ast }$ of $u\in W^{1,p}({\mathbb{R}}^n)$ has a classical Fréchet derivative at almost every point, equal to its weak derivative. At a point $x$ where the $p$-mean oscillation of $Du$ tends to zero, apply the Morrey inequality on a cube to $u^{\ast }(y)-u^{\ast }(x)-Du(x)\cdot (y-x)$. On a cube of side comparable to $|y-x|$, its oscillation is at mostThe required points have full measure by the Lebesgue differentiation theorem.

Define $f(x,y)=x^3/(x^2+y^2)$ away from the origin and zero at the origin. Every directional derivative exists. Along a unit vector $n$, the derivative at the origin is $n_1^3$, which is not linear in $n$. Thus these directional derivatives cannot be the restrictions of one Fréchet derivative, and the function is not Fréchet differentiable at that point.

Suppose a perturbation $\delta x$ satisfies $\Vert \delta x\Vert \le \epsilon$, with $\epsilon \ge 0$. Absolute homogeneity of a norm and the dual norm definition giveFor nonzero $\delta x$, divide it by its norm to obtain an input in the unit ball; for zero perturbation the inequality is immediate. This is an exact error bound for a linear functional, for every perturbation in the stated input set.

For $p(x)=2x_1-x_2$ and a budget $\epsilon =0.1$, the sharp output-error bounds are $0.2$ for $\Vert \delta x{\Vert }_1\le 0.1$, $\sqrt{5}/10\simeq 0.224$ for $\Vert \delta x{\Vert }_2\le 0.1$, and $0.3$ for $\Vert \delta x{\Vert }_{\infty }\le 0.1$. Scale the maximizing vectors from the preceding example by $0.1$ to attain the positive endpoints; their negatives attain the negative endpoints. The L1 norm budget constrains a total absolute error; the maximum norm budget permits each coordinate to have that error separately. These budgets define nested sets and carry different information.

The coordinates must use the stated scale and units. Rescaling one coordinate changes both its coefficient in $a$ and the meaning of its input budget. For a nonlinear differentiable function with scalar values, its Fréchet derivative at the base point is a linear functional governing the first-order response. An exact finite-perturbation error bound additionally needs control of the Taylor remainder or of the Fréchet derivative along the path.

For a positive integer $m$, the negative matrix power is differentiable on the open general linear group. The Neumann series gives $(A+H{)}^{-1}=A^{-1}-A^{-1}H{A}^{-1}+O(\Vert H{\Vert }^2)$ locally around any invertible matrix. Applying the product rule to the $m$ factors of this inverse gives the displayed Fréchet derivative. Each summand chooses one inverse factor to differentiate while keeping the others in order.

The factor order matters. Only when $A$ and $H$ commute does the sum simplify to $-m{A}^{-(m+1)}H$. This is a useful example of why scalar differentiation rules for powers cannot be transferred to matrices without accounting for noncommutativity.

For $I(u)={\int }_a^{b}f(x,u,u^{\prime })dx$ on $W^{1,\infty }$, continuity of $f,f_u,f_v$ on $[a,b]\times {\mathbb{R}}^2$ gives a bounded Fréchet derivative in the displayed form. Fixing $u$ and restricting $\Vert w{\Vert }_{W^{1,\infty }}$ to a bounded neighborhood confines all intermediate pairs $(u+tw,u^{\prime }+t{w}^{\prime })$ to a compact range up to null sets. Uniform continuity of the partial derivatives bounds the integrated first-order remainder by $(b-a)\omega (\Vert w\Vert )(\Vert w{\Vert }_{\infty }+\Vert w^{\prime }{\Vert }_{\infty })$, where $\omega (s)\to 0$. This is little-oh of the Sobolev norm. The result needs no continuous derivative representative for $u$; essential boundedness suffices.

The inequalities $0\le \log \cosh t\le t^2/2$ make this energy finite on $H^1$. Since the second derivative of its scalar potential is ${\mathrm{sech}}^{2}t\in (0,1]$, its Fréchet derivative is $DI(v)[h]=\int (\nabla v\cdot \nabla h+\tanh (v)h)$, with Taylor remainder between zero and half the squared $H^1$ norm of $h$. The derivative operator is strictly monotone because the hyperbolic tangent is strictly increasing.

This convex function penalizes positive densities using negative entropy, with $0\log 0=0$ and infinite value for negative inputs. Its scalar derivative is $\log x$ on $x>0$. On a bounded domain, restricting to uniformly positive bounded functions gives a simple setting for a Fréchet derivative. Its Bregman distance is the generalized Kullback–Leibler divergence. With fixed total mass, minimizing this negative-entropy expression maximizes the corresponding entropy.

Let $B=\{p:\Vert p-p_0\Vert \le r\}$ and write $\epsilon =\Vert A(p_0)-p_0\Vert$. Integrating the Fréchet derivative along the line segment between two points in the convex set givesand the operator norm bound impliesIn particular, for $p\in B$,Thus the a posteriori contraction ball is mapped into itself. The ball is a closed subset of a Banach space, hence is a complete metric space. Applying the contraction mapping theorem to $A:B\to B$ gives existence and uniqueness:The estimate also gives $\Vert p_{\ast }-p_0\Vert \le \epsilon /(1-\kappa )$, by applying the same triangle inequality to $A(p_{\ast })=p_{\ast }$. Equality in the residual bound is allowed; strict contraction mapping comes from $\kappa <1$, and a fixed point on the closed ball's boundary is included.

By equivalent norms, choose $c,C>0$ with $c\Vert h{\Vert }_1\le \Vert h{\Vert }_2\le C\Vert h{\Vert }_1$. Differentiability in the first norm means that there is a bounded linear map $L:V\to \mathbb{R}$ with $f(a+h)-f(a)-Lh=r(h)$ and $|r(h)|/\Vert h{\Vert }_1\to 0$. The same $L$ is bounded in the second norm, since $|Lh|\le K\Vert h{\Vert }_1\le K{c}^{-1}\Vert h{\Vert }_2$. Moreover,The two norms define the same approach $h\to 0$, so this proves differentiability with the same Fréchet derivative in the second norm. Reversing the norms proves the converse.

For completeness, the derivative is unique: if $L_1,L_2$ both work, fix $v$ and take $h=tv$. The two remainder estimates imply $|(L_1-L_2)v|=o(1)$ as $t\to 0$, so $(L_1-L_2)v=0$ for every $v$. Thus differentiability and its derivative are independent of the chosen norm. This is differentiability under equivalent norms, not merely invariance of continuity.

Continuity on the compact product of unit spheres bounds the bilinear map: there is $C$ such that $\Vert f(h,k){\Vert }_3\le C\Vert h{\Vert }_1\Vert k{\Vert }_2$. This follows by scaling each nonzero vector to unit norm; if one factor space is zero, the map and the relevant assertion are trivial.

Bilinearity gives the exact expansionThe proposed derivative $L(h,k)=f(h,b)+f(a,k)$ is linear and bounded, with $\Vert L(h,k){\Vert }_3\le C(\Vert b{\Vert }_2\Vert h{\Vert }_1+\Vert a{\Vert }_1\Vert k{\Vert }_2)$. Set $\rho =(\Vert h{\Vert }_1^2+\Vert k{\Vert }_2^2{)}^{1/2}$. Its remainder obeysThus the derivative of a continuous bilinear map is the Fréchet derivative

Differentiability at $p$ means there is a linear map $A=Df{|}_p$ such thatThis is the Fréchet derivative definition; the same linear approximation must work in every direction, not merely along coordinate axes.

The chain rule says that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, then $g\circ f$ is differentiable at $p$ andFor its proof, put $A=Df{|}_p$, $B=Dg{|}_{f(p)}$, and $k=f(p+h)-f(p)=Ah+r(h)$. Since a linear map between these finite-dimensional normed spaces is bounded, $\Vert k\Vert =O(\Vert h\Vert )$. Differentiability of $g$ gives $g(f(p)+k)=g(f(p))+Bk+s(k)$ with $s(k)=o(\Vert k\Vert )$. ThereforeIf $k=0$, take $s(0)=0$; otherwise $\Vert s(k)\Vert /\Vert h\Vert =(\Vert s(k)\Vert /\Vert k\Vert )(\Vert k\Vert /\Vert h\Vert )\to 0$. This proves the required full linear approximation.

For the Euclidean norm, at every $x\ne 0$ the ordinary one-coordinate differentiation givesAt the origin its difference quotient in direction $e_i$ is $|t|/t$, whose right and left limits are $1$ and $-1$. Thus none of its coordinate partial derivatives exists at the origin.

Work with real functions and $f\in L^2(0,1)$. To give the variational problem a definite domain, choose the homogeneous Dirichlet boundary conditions used in the next part. The natural form space is the zero-trace Sobolev space $H=H_0^1(0,1)$, with norm $\Vert v{\Vert }_H=\Vert v^{\prime }{\Vert }_2$. The Sobolev trace operator is well defined because one-dimensional $H^1$ functions have absolutely continuous representatives. For example $v(0)={\int }_0^{1}v(x)dx-{\int }_0^1(1-x)v^{\prime }(x)dx$, so its endpoint trace is bounded in the $H^1$ norm; the other endpoint has the analogous estimate. The zero-trace space is closed in the Hilbert space $H^1$, and the Poincaré inequality $\Vert v{\Vert }_2\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_2$ makes the chosen derivative norm equivalent to its full norm. Thus $H$ is a Hilbert space.

Differentiability of $p,q$ implies continuity on the compact interval. Hence there are finite constants $p_{\max },q_{\max }$ and a positive $p_{\min }$ with $p_{\min }\le p\le p_{\max }$ and $0\le q\le q_{\max }$. Define the symmetric bilinear formThe Cauchy-Schwarz inequality and Poincaré inequality give all the estimates required by the Lax-Milgram theorem:That theorem states that a continuous coercive bilinear form on a Hilbert space and a bounded linear functional determine a unique weak solution. Completeness, continuity, coercivity and boundedness have now all been verified. Therefore there is a unique $u\in H$ with $a(u,v)=⟨f,v⟩$ for every $v\in H$.

The energy version of the given variational problem is $I(v)=a(v,v)-2⟨f,v⟩$. Its exact variation isBoundedness of $a$ makes the last term quadratic in $\Vert w{\Vert }_H$, so the Fréchet derivative is $DI(v)[w]=2a(v,w)-2⟨f,w⟩$. Its vanishing for every test function is preciselyTaking compactly supported smooth tests and integrating in the distributional sense shows that this is the Euler-Lagrange equation $-(p{u}^{\prime }{)}^{\prime }+qu=f$. Conversely that weak equation gives the displayed stationary condition. The stationary point is actually the unique global minimizer, becauseEquality occurs only when $v=u$. This proves the variational equivalence, including existence and uniqueness rather than only a formal derivative calculation.

For precision about the literal L2 inner product $⟨Lv,v⟩$, use the flux-domain realization of a one-dimensional elliptic operatorFor such $v$, weak integration by parts and its zero endpoint values give $⟨Lv,v⟩=a(v,v)$. On the larger form space $H$, the same energy is interpreted by the weak operator pairing. The minimizer belongs to $D(L)$ automatically: its weak equation gives $(p{u}^{\prime }{)}^{\prime }=qu-f\in L^2$, while $p{u}^{\prime }\in L^2$. Thus its equation also holds as an equality of $L^2$ functions. This domain does not assume bounded $p^{\prime }$, which mere differentiability would not supply. If $p\in C^1$ one may equivalently use $D(L)=H^2\cap H_0^1$, but that stronger coefficient assumption is not needed for the proof here.

Let $p=(x_0,y_0)$, $a=f_x(p)$ and $b=f_y(p)$. For sufficiently small $h,k$, decompose the increment into an $x$-segment at height $y_0+k$ and a $y$-segment at $x_0$. The one-variable mean value theorem applies on each segment because existence of its partial derivative gives continuity along the segment. Thus, when both increments are nonzero,for some $0<\theta ,\eta <1$. The same formula with the absent segment omitted covers zero increments. Continuity of both partial derivatives at $p$ makes the remainder after subtracting $ah+bk$ at most $\epsilon (|h|+|k|)$ for small increments, and hence at most $\sqrt{2}\epsilon \sqrt{h^2+k^2}$. Therefore the Fréchet derivative is $(h,k)\mapsto ah+bk$. Continuity throughout the disc is not required; continuity at the point suffices.

For the determinant, expand by multilinearity of the determinant in the columns of $I+H$. Taking no perturbed column gives one; taking exactly one gives $\mathrm{tr}H$; all other terms have degree at least two in the entries of $H$. Since matrix-entry functionals are bounded in any finite-dimensional norm, their sum is $O(\Vert H{\Vert }^2)$. HenceFor invertible $A$, multiplicativity of the determinant gives $\det (A+H)=\det A\det (I+A^{-1}H)$. The bounded linear map $H\mapsto A^{-1}H$ transfers the preceding remainder estimate, giving the derivative of the determinantTo justify an open invertible neighborhood, let $\Vert K\Vert <1$. Completeness and submultiplicativity make the Neumann series $S={\sum }_{j=0}^{\infty }(-K{)}^j$ converge in matrix norm. Multiplying finite partial sums by $I+K$ gives $I-(-K{)}^{N+1}$, which tends to $I$. Continuity of multiplication proves $(I+K)S=S(I+K)=I$. Thus $I+K$ is invertible, and the series gives $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$ near zero.

Apply the first-derivative formula at $I+H$ to a test matrix $K$:The last estimate is uniform for $\Vert K\Vert \le 1$, so it proves differentiability of the derivative in the operator norm, rather than merely a directional second derivative. The second derivative of the determinant is the symmetric bilinear mapIts symmetry follows from $\mathrm{tr}(HK)=\mathrm{tr}(KH)$.

Differentiability means existence of a linear map $A:{\mathbb{R}}^2\to \mathbb{R}$ such thatThis is the Fréchet derivative. Substituting increments on the coordinate axes shows $A(h,k)=f_x(x_0,y_0)h+f_y(x_0,y_0)k$. For the unit direction $v=(\cos \alpha ,\sin \alpha )$, the remainder is $r(tv)=o(|t|)$. Dividing the increment by $t$ proves differentiability of the restricted function and gives its directional derivative:For the supplied rational function, both coordinate-axis restrictions are zero, so $f_x(0,0)=f_y(0,0)=0$. Any Fréchet derivative would therefore be the zero map. But $f(t,t)=t$, and hence $|f(t,t)|/\sqrt{t^2+t^2}=1/\sqrt{2}$ for nonzero $t$. This contradicts the required vanishing remainder. ThusIn fact every directional derivative exists: along $v$ it equals ${\cos }^2\alpha \sin \alpha +\cos \alpha {\sin }^2\alpha$. These values do not depend linearly on the direction, illustrating why existence of directional derivatives is weaker than differentiability.

The Fréchet derivative at $p=(x_0,y_0)$ is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ for whichThis is differentiability; if it holds, $L(h_1,h_2)=D_{1}f(p)h_1+D_{2}f(p)h_2$.

To prove the sufficient condition, take a rectangle about $p$ contained in the open domain. Split the increment along its coordinate directions:The one-variable mean value theorem, applicable because the restrictions are differentiable, represents these two terms asThe zero-increment cases are interpreted directly. Subtract $D_{1}f(p)h+D_{2}f(p)k$. By continuity of both partial derivatives at $p$, the remainder is at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ once the increment is small. This is the required little-oh remainder and proves differentiability.

The converse is false. Set $f(x,y)=x^2\sin (1/x)$ for $x\ne 0$, and $f(0,y)=0$. At each point with $x=0$, its Fréchet derivative is zero, since the function increment has magnitude at most $h^2=o(\sqrt{h^2+k^2})$. It is smooth away from this line. Butis not continuous there. Thus a function can be differentiable everywhere while its partial derivatives are discontinuous.

For $h$, every point with $x\ne 0$ is a smooth point. At the origin, $|h(x,y)|\le |xy|\le (x^2+y^2)/2$, so $h$ is differentiable with zero derivative. At $(0,y_0)$ with $y_0\ne 0$, the putative first partial derivative would require the limit $y_0\sin (1/x)$, which does not exist. Consequently the exact set of differentiability is

Put $r=\Vert X\Vert <1$. By submultiplicativity of the operator norm, $\Vert X^k\Vert \le r^k$, so the Neumann series converges absolutely in the complete finite-dimensional matrix space. Its partial sums satisfyTaking limits gives the inverse and a quantitative remainder bound:To establish two actual Fréchet derivatives, use the resolvent identityFor fixed $X\in U$ and sufficiently small $H$, factor $I-X-H=(I-X)(I-f(X)H)$ and apply the Neumann series to the second factor. This proves that $X+H\in U$, that $f(X+H)$ is locally bounded, and that $f(X+H)-f(X)=O(\Vert H\Vert )$. Subtracting $f(X)Hf(X)$ in the identity leaves $O(\Vert H{\Vert }^2)$. Thus $Df(X)[H]=f(X)Hf(X)$ is the Fréchet derivative.

Near zero, $f(X)=I+X+O(\Vert X{\Vert }^2)$, uniformly in the operator norm. Hence, as a linear operator in $H$,This proves differentiability of $Df$ at zero, givingThe order of multiplication matters: the second Fréchet derivative is the symmetric bilinear map $HK+KH$, not $2HK$.

Differentiability here is understood in the sense of the Fréchet derivative: there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe map $L=Df(x)$ is the Fréchet derivative. In finite dimensions a linear map is bounded, so $\Vert Lh\Vert \le C\Vert h\Vert$ for some $C$. The defining estimate therefore impliesThis is precisely continuity at $x$: differentiability implies continuity.