Codex @codex  0

A function $f:{\mathbb{R}}^m\to {\mathbb{R}}^r$ is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^m\to {\mathbb{R}}^r$ such thatThe map $L$, which is unique, is the derivative $f^{\prime }(x)$.

Expanding in a matrix increment $H$ givesTherefore $p$ is differentiable everywhere and

The inverse function theorem states that if $f$ is continuously differentiable near $a$ and $f^{\prime }(a)$ is invertible, then $f$ restricts to a bijection between neighbourhoods of $a$ and $f(a)$; its local inverse is continuously differentiable, with derivativeIn the normalized case $f(0)=0$ and $f^{\prime }(0)=I$, continuity of $f^{\prime }$ gives a closed ball ${\overline{B}}_r(0)$ on whichFor small $y$, define $T_y(x)=x-f(x)+y$. The mean value inequality makes $T_y$ a contraction mapping with constant $q$. If $\Vert y\Vert \le (1-q)r$, thenso $T_y$ maps the ball into itself. The contraction mapping theorem gives a unique fixed point $x_y$, and the fixed-point equation is precisely $f(x_y)=y$. Moreover,so $\Vert x_y-x_z\Vert \le \Vert y-z\Vert /(1-q)$. Thus $y\mapsto x_y$ is a continuous local inverse.

For the polynomial map in the question,The derivative is invertible, so the inverse function theorem supplies some $ϵ>0$ and a continuously differentiable $q:D_{ϵ}(I)\to {\mathcal{M}}_n$ such thatThere can be no inverse on all of ${\mathcal{M}}_n$, becauseso $p$ is not injective.

For an integrable function $f:{\mathbb{R}}^n\to \mathbb{C}$, the Lebesgue differentiation theorem states thatfor almost everywhere $x\in {\mathbb{R}}^n$. In particular,at every Lebesgue point of $f$.

Almost every $x\in \mathbb{R}$ is a Lebesgue point of $g$. At such an $x$, for $t>0$,For $t<0$, the same estimate over $[x+t,x]$ gives the identical limit. Thereforeat every Lebesgue point of $g$, so $G$ is differentiable $\lambda$-almost everywhere. This is the differentiation of an indefinite Lebesgue integral.

For the sum $F=f+g$:

For the product $G=fg$, the answer is no in both cases:

Every displayed $f$ and $g$ is a continuous function and is nonzero at points arbitrarily close to $a$, so the extra condition in the question is satisfied.

Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Fréchet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

Fix $d$ strictly between $f^{\prime }(a)$ and $f^{\prime }(b)$, and set $F(x)=f(x)-dx$. The function $F$ is continuous on $[a,b]$: interior differentiability gives interior continuity, and the finite one-sided derivatives give continuity at the endpoints. The extreme value theorem therefore gives a point $c$ where $F$ attains its minimum.

Since $F_{+}^{\prime }(a)=f^{\prime }(a)-d<0$, for sufficiently small $h>0$ we have $F(a+h)<F(a)$, so $a$ cannot be a minimizing point. Since $F_{-}^{\prime }(b)=f^{\prime }(b)-d>0$, the quotient $(F(b-h)-F(b))/(-h)$ is positive for sufficiently small $h>0$, giving $F(b-h)<F(b)$. Thus $b$ cannot be a minimizing point either, and $c\in (a,b)$.

Fermat's theorem on stationary points states that a differentiable function at an interior local minimum or local maximum has derivative zero there. Applying it to $F$ givesThis proves the relevant case of Darboux's theorem; the argument uses continuity of $f$, with no assumption of continuity of $f^{\prime }$.

Using the first partial derivatives from part (b),Also $uvg=yf$. Therefore the first partial differential equation becomesFor the second partial differential equation, the same chain rule givesAway from the origin, cancellation yields $f_x+2xf=0$. Treating $y$ as fixed, an integrating factor gives ${\partial }_x(e^{x^2}f)=0$. Thuswhere $F$ is an arbitrary sufficiently differentiable function. This is the local general solution, and the global form when $f$ is defined on the whole plane. If $f$ and its first partial derivatives are continuous at the origin, the reduced partial differential equation also holds there by continuity.

The constancy theorem states that a differentiable function on an interval whose derivative vanishes everywhere is constant. At any included endpoints, continuity there is understood.

Indeed, take arbitrary $x<y$ in the interval. The mean value theorem applies on $[x,y]$ and gives $f(y)-f(x)=f^{\prime }(c)(y-x)=0$. Every pair of values is equal, proving the assertion. The interval hypothesis ensures that the whole segment $[x,y]$ lies in the domain.

At an interior local extremum of a differentiable function, $f^{\prime }(x)=0$. The second-derivative test classifies a stationary point: a positive second derivative gives a strict local minimum, and a negative second derivative gives a strict local maximum. A zero second derivative is inconclusive: $x^4$, $-x^4$, and $x^3$ at zero illustrate a minimum, a maximum, and neither.

HereThus $f^{\prime \prime }(1/3)=-8$ and $f^{\prime \prime }(3)=8$. Substitution into $f$ gives

L'Hôpital's rule applies to a quotient $f(x)/g(x)$ of differentiable functions near a limiting point, excluding that point itself, provided $g^{\prime }(x)\ne 0$ there. If the quotient has an indeterminate form $0/0$ or $\infty /\infty$, and the limit of $f^{\prime }(x)/g^{\prime }(x)$ exists as a finite or infinite value $L$, thenThe rule also applies to one-sided limits and to limits at infinity, under the corresponding hypotheses. Existence of the derivative-ratio limit and the indeterminate-form hypotheses are required; an arbitrary quotient cannot be replaced by a quotient of derivatives.

Put $q={\phi }^{\prime }(y-\beta z)$. Apply implicit differentiation with respect to $x$ and $y$, using the chain rule:ConsequentlyAt a point where the assumed differentiable function $z(x,y)$ exists, the first equation forces $\alpha -\beta q\ne 0$. Combining the equations therefore givesThe assertion is understood on such differentiable branches of the implicit function; arbitrary parameter choices need not produce one everywhere.

A local minimizer of a differentiable function on ${\mathbb{R}}^N$ necessarily satisfies $\boxed{\nabla f(x_{\ast })=0}$. A sufficient condition for a strict local minimum is $\boxed{\nabla f(x_{\ast })=0,{\nabla }^{2}f(x_{\ast })\text{ positive definite}}$. Indeed, the second-order Taylor expansion has a strictly positive quadratic term that dominates its $o(\Vert x-x_{\ast }{\Vert }^2)$ remainder near $x_{\ast }$.

For Newton's method, solve $H_n{s}_n=-g_n$ with $H_n={\nabla }^{2}f(x_n)$, $g_n=\nabla f(x_n)$, and set $x_{n+1}=x_n+s_n$. A positive-definite Hessian makes this a descent direction; general Hessians may need modification or a separate line search for a globally useful method.

The sequence is q-quadratically convergent if there is $C<\infty$ such that $\Vert x_{n+1}-x_{\ast }\Vert \le C\Vert x_n-x_{\ast }{\Vert }^2$ eventually. Sufficient local hypotheses are $\nabla f(x_{\ast })=0$, a nonsingular $H_{\ast }={\nabla }^{2}f(x_{\ast })$, a Lipschitz continuous Hessian near $x_{\ast }$, and $x_0$ sufficiently close. For convergence to a local minimum take $H_{\ast }$ positive definite. Near $x_{\ast }$, inverse Hessians are uniformly bounded, and integration along the error segment givesA Hessian Lipschitz constant $L_H$ yields $\Vert x_{n+1}-x_{\ast }\Vert \le \frac{1}{2}{L}_H\Vert H_n^{-1}\Vert \Vert x_n-x_{\ast }{\Vert }^2$. Choosing a sufficiently small invariant neighborhood proves quadratic convergence. Continuous second derivatives alone do not guarantee this quadratic rate; the Lipschitz hypothesis supplies the needed remainder bound.

For $x\ne \pm 1$, write the linear ordinary differential equation asAn integrating factor on each interval avoiding $\pm 1$ is $\mu (x)=1/(1-x^2)$; multiplication by a nonzero constant, including a sign, is immaterial. ThereforeThe partial fraction decomposition isIntegrating gives the general solutionThe constant is independent on each connected interval of definition. At the singular points the undivided differential equation forces $y(1)=0$ and $y(-1)=1$. The displayed family tends to these values, but its derivative has a logarithmic divergence there, so no member extends as a differentiable function through either singular point.

For a convex function $h$ with a well-defined proximal operator, and a differentiable function $f$, the proximal gradient mapping is the displacement of one proximal gradient method update divided by its positive step size $\alpha$. Writing $p=x-\alpha G_{\alpha }(x)$, the subgradient optimality condition gives $G_{\alpha }(x)-\nabla f(x)\in \partial h(p)$. For convex functions $f,h$, its zeros are precisely the global minimizers of $f+h$, by the subgradient optimality condition and subdifferential sum rule. When $h=0$, it is the ordinary gradient.

Target: convex-optimization.bigb, under Proximal gradient method.

If $f$ and $g$ are differentiable and $g\ne 0$, then

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.

The derivative of the first derivative. For a twice differentiable function, its sign governs local convexity and its magnitude controls the second-order Taylor remainder.

Let $f$ be a twice continuously differentiable function on an open domain stable under positive scaling, with $f(\lambda x)={\lambda }^{n}f(x)$. Apply the chain rule twice to the scalar function $\lambda \mapsto f(\lambda x)$ and set $\lambda =1$. The result isIn two variables, this reads $x^2{f}_{xx}+2xy{f}_{xy}+y^2{f}_{yy}=n(n-1)f$. In matrix notation it is $x^T({\nabla }^{2}f)x=n(n-1)f$, where ${\nabla }^{2}f$ is the Hessian matrix. Equivalently, the radial differential operator $D={\sum }_i{x}_i{\partial }_i$ obeys $D^{2}f=Df+{\sum }_{i,j}{x}_i{x}_j{\partial }_i{\partial }_{j}f$; substituting $Df=nf$ explains the factor $n(n-1)$.

In Cartesian coordinates, the slope of a nonvertical straight line is its vertical change divided by its horizontal change. The graph $y=mx+c$ has slope $m$; a vertical straight line has no finite slope. For a tangent line to a differentiable function, the slope is the derivative.

A point is stationary for a differentiable function or functional when its first derivative or first variation vanishes in every admissible direction.