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An antiderivative of $g$ is a differentiable function $p$ satisfying $p^{\prime }=g$.

Let $f$ be a differentiable function on an open domain stable under positive rescaling, with $f(\lambda x)={\lambda }^{k}f(x)$ for $\lambda >0$. The chain rule differentiates this homogeneous function identity at $\lambda =1$ to giveIf $f$ is twice differentiable, differentiating once more gives ${\sum }_i{x}_i{\partial }_i{\partial }_{j}f=(k-1){\partial }_{j}f$. In degree one its Hessian matrix therefore annihilates the radial vector $x$. This is the identity used in the no-scale identity from degree-one homogeneity. For an extensive quantity, the degree-one formula also supplies the thermodynamic identity behind the Gibbs-Duhem equation.

At an interior local extremum of a differentiable function, its gradient is zero: restricting the function to every line through the point shows that all directional derivatives vanish. This is only a necessary condition in general. For a strictly concave function, an interior stationary point is the unique global maximizer on its convex domain. Constrained or boundary optima instead need conditions adapted to their feasible directions.

If $A(t)=A+tE$ and $\lambda (t)$ is a differentiable function defining a simple eigenvalue, thenThe Cauchy-Schwarz inequality and the operator norm give

The gradient of a scalar-valued differentiable function is the vector of its first partial derivatives.

A function is homogeneous of degree $k$ when rescaling its arguments by a common scalar $\lambda$ rescales its value by ${\lambda }^k$, whenever the arguments and that power are defined. For integer $k$ over a vector space, one can require every nonzero scalar; a homogeneous polynomial gives an example. On a domain stable under positive real rescaling, the positive-scaling version allows real $k$. Degree-one positive homogeneity is the convention used for norms and many convex functions. The Euler theorem for homogeneous functions turns this scaling law into a differential identity for a differentiable function.

Every real-valued monotone function on a real interval is differentiable with a finite derivative at almost every point.

A stationary point is nonisolated if every neighbourhood contains other stationary points. For a twice differentiable function, a regular smooth curve of stationary points through a point forces its Hessian matrix to annihilate the curve's nonzero tangent: differentiate the vanishing gradient along the curve. Thus the Hessian matrix has a zero eigenvalue there, but that alone need not determine whether the point is a local minimum or local maximum.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. We first use it to show that a continuous function on $[a,b]$ is bounded above. Otherwise, choose $x_n\in [a,b]$ with $f(x_n)>n$. A convergent subsequence $x_{n_k}\to c$ exists, with $c\in [a,b]$ because the interval is closed. Continuity would give $f(x_{n_k})\to f(c)$, contradicting $f(x_{n_k})>n_k\to \infty$.

Let $L=\sup \{f(x):x\in [a,b]\}$, which is finite. By the definition of supremum, there are $y_n\in [a,b]$ with $L-1/n<f(y_n)\le L$. Another application of the Bolzano-Weierstrass theorem gives a subsequence $y_{n_k}\to c\in [a,b]$. By continuity,Thus the supremum is actually attained:This proves the maximum part of the extreme value theorem from sequential compactness rather than assuming it.

Now suppose $f^{\prime \prime }<0$ on $\mathbb{R}$. Applying the mean value theorem to $f^{\prime }$ shows that $f^{\prime }$ is strictly decreasing. At a local maximum of a differentiable function, the left difference quotients are nonnegative and the right difference quotients are nonpositive; existence of the derivative forces $f^{\prime }=0$. A strictly decreasing function has at most one zero. ThereforeIf that zero exists, $f^{\prime }$ is positive to its left and negative to its right, so the local maximum is also a global maximum.

For the uniform bound by $K<0$, defineIts derivative is $h^{\prime }(x)=f^{\prime \prime }(x)-K<0$, so $h$ is strictly decreasing by the mean value theorem. Comparing with $h(0)=f^{\prime }(0)$ givesBecause $K<0$, this makes $f^{\prime }$ negative at sufficiently large positive $x$ and positive at sufficiently large negative $x$. The function $f^{\prime }$ is continuous, since it is differentiable; the intermediate value theorem supplies a zero $c$ between such points. Its strict decrease makes this zero unique. Again using the mean value theorem for $f$, the derivative signs imply that $f$ increases up to $c$ and decreases after $c$:This illustrates uniform negative curvature and global maximization.

Merely requiring $f^{\prime \prime }<0$ does not force existence. TakeThen $f^{\prime }(x)=e^{-x}>0$ and $f^{\prime \prime }(x)=-e^{-x}<0$, but $f$ is strictly increasing, approaches $0$ as $x\to +\infty$, and never equals $0$. The answer to the final question is no, even for a function bounded above:The stronger uniform curvature assumption prevents this behaviour. Here $f^{\prime \prime }(x)\to 0$ at the end where the supremum is approached, so no uniform negative bound of that kind is available.

Choose $X=x-2y$ and $T=x$, so $(\alpha ,\beta ,\gamma ,\delta )=(1,-2,1,0)$. The Jacobian determinant of the coordinate change is two, so it is an invertible linear map. By the chain rule, ${\partial }_x={\partial }_X+{\partial }_T$ and ${\partial }_y=-2{\partial }_X$. ConsequentlyThis is the wave equation with $c=1$. The D'Alembert formula is $F(X,T)=U(X+T)+V(X-T)$, with arbitrary twice differentiable functions $U,V$. Returning to the old coordinates and absorbing factors of two into the arbitrary functions givesDirect differentiation checks the cancellation. Equivalently, first integrating ${\partial }_x(f_x+f_y)=0$ produces the same two-function family.

Differentiability at $x_0$ meansHence $f(x)-f(x_0)=(x-x_0)(f^{\prime }(x_0)+o(1))\to 0$ as $x\to x_0$. Therefore every differentiable function is continuous at each point where it is differentiable.

The map $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ when there is a linear map $Df(a)$ such thatIt is $C^1$ when it is differentiable everywhere and $a\mapsto Df(a)$ is continuous. The definition immediately gives $\Vert f(a+h)-f(a)\Vert \le (\Vert Df(a)\Vert +o(1))\Vert h\Vert \to 0$, so differentiability implies continuity.

The inverse function theorem says that if $f$ is $C^1$ near $a$ and $Df(a)$ is invertible, then $f$ restricts to a diffeomorphism between neighbourhoods of $a$ and $f(a)$.

No. DefineEach $f_n$ is a differentiable function on $[-1,1]$, whileThus $f_n$ converges uniformly to the absolute value function $|x|$, which is not differentiable at $0$. Uniform convergence alone therefore does not preserve differentiability.

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 Frechet 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

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

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.