Differentiability at a point and one-sided derivatives

$y=\sqrt{1-e^{-a^{2}x^{2}}}$ at the point $x=0$,

$y=\arcsin \dfrac {2x}{1+x^{2}}$ at the point $x=1$

Prove: if the function $f(x)$: (1) is defined and continuous on the closed interval $[x_{0},X]$; (2) has a finite derivative $f'(x)$ on the interval $(x_{0},X)$; (3) has a finite or infinite limit

then the finite or infinite one-sided derivative $f'_{+}(x_{0})$ exists, and

Prove that the function

has a finite limit

but the function $f(x)$ has neither one-sided derivative $f'_{-}(1)$ nor $f'_{+}(1)$.

Give a geometric illustration of this fact.

it is continuous;

it is differentiable at the irrational points and not differentiable at the rational points.

if a function has a derivative at some point, then it is continuous at that point;

if a function is continuous at some point, then it has a derivative at that point.

is continuous;

has a derivative;

have a continuous derivative?

is continuous;

has a derivative;

have a continuous derivative?

$y = \begin{cases} \alpha x+\beta , & \text{ if } x \leqslant 1, \\ x^{2}, & \text{ if } x > 1 ;\end{cases}$

$y = \begin{cases} \alpha +\beta x^{2}, & \text{ if }\left|x\right| \leqslant 1, \\ 1 /\left|x\right|, & \text{ if }\left|x\right| > 1 ; \end{cases}$

$y = \begin{cases} \alpha x^{3}+\beta x, & \left|x\right| \leqslant 2, \\ (1 / \pi ) \arcsin (1 / x), & \left|x\right| > 2 \end{cases}$

$y = \begin{cases} 2 x-2, & \text{ if } & x \leqslant 1, \\ \alpha (x-1)(x-2)(x-\beta ), & \text{ if } & 1 < x < 2, \\ x / 2-1, & \text{ if } & x \geqslant 2. \end{cases}$

$y = \begin{cases} (x+\alpha ) e^{-\beta x}, & \text{ if } x < 0, \\ \alpha x^{2}+\beta x+1, & \text{ if } x \geqslant 0 ; \end{cases}$

$y = \begin{cases} \alpha x+\beta , & \text{ if } x < 0, \\ \alpha \cos x+\beta \sin x, & \text{ if } x \geqslant 0. \end{cases}$

at the point $x = 1$;

at the point $x = -1$.

$y = \left|x^{3}(x+1)^{2}(x+2)\right|$;

$y = \left|\sin x\right|$

$y = x\left|x\right|$;

$y = \left|\pi -x\right| \sin x$

$y = \arccos (\cos x)$

$y = \begin{cases} x^{3}, & \text{ if } x \leqslant 0 \\ e^{-1 / x}, & \text{ if } x > 0 \end{cases}$

$y = \begin{cases} x^{2}\left|\cos (\pi / x)\right|, & \text{ if } x \neq 0, \\ 0, & \text{ if } x = 0 ; \end{cases}$

$y = \begin{cases} x, & \text{ if } x-\text{ rational number, } \\ 0, & \text{ if } x-\text{ irrational number. }\end{cases}$

$y = \begin{cases} x^{2}, & \text{ if } x \text{ — rational number, } \\ 0, & \text{ if } x \text{ — irrational number. } \end{cases}$

$y = \begin{cases} x^{2}, & \text{ if } x \text{ — rational number, } \\ 2\left|x\right|-1, & \text{ if } x \text{ — irrational number. } \end{cases}$

if a function $f$ has a derivative at some point, and a function $g$ does not, then the function $f+g$ does not have a derivative at that point;

if the functions $f$ and $g$ do not have a derivative at some point, then the function $f+g$ also does not have a derivative at that point;

if a function $f$ has a derivative at some point, and a function $g$ does not, then the function $fg$ also does not have a derivative at that point;

if the functions $f$ and $g$ do not have a derivative at some point, then the function $fg$ also does not have a derivative at that point.

$f^{\prime }\left(g\left(x_{0}\right)\right)$ exists, $g^{\prime }\left(x_{0}\right)$ does not exist;

$f^{\prime }\left(g\left(x_{0}\right)\right)$ does not exist, $g^{\prime }\left(x_{0}\right)$ exists;

$f^{\prime }\left(g\left(x_{0}\right)\right)$ and $g^{\prime }\left(x_{0}\right)$ do not exist.

if for functions $f$ and $g$ differentiable on the interval $(a ; b)$ the inequality $f < g$ holds, then $f^{\prime } \leqslant g^{\prime }$ on $(a ; b)$;

if on the interval $(a ; b)$ the inequality $f^{\prime } < g^{\prime }$ holds, then $f < g$ on $(a ; b)$;

if $f(a) = g(a)$ and $f^{\prime }(x) < g^{\prime }(x)$ on the interval $(a ; b)$, then $f(x) < g(x)$ on $(a ; b)$.

for a differentiable function $y(x), x \in (a ; b)$, to have a derivative that is monotonic on the interval $(a ; b)$, it is necessary that $y(x)$ be monotonic on $(a ; b)$;

for a differentiable function $y(x), x \in (a ; b)$, to have a derivative that is monotonic on the interval $(a ; b)$, it is sufficient that $y(x)$ be monotonic on the interval $(a ; b)$;

for a differentiable function to have a periodic derivative, it is necessary that the function be periodic;

for a differentiable function to have a periodic derivative, it is sufficient that the function be periodic.

for the derivative of a differentiable function to be an even function, it is sufficient that the function be odd;

for the derivative of a differentiable function to be an even function, it is necessary that the function be odd;

for the derivative of a differentiable function to be an odd function, it is necessary that the function be even;

for the derivative of a differentiable function to be an odd function, it is sufficient that the function be even.

if the function $y(x)$ is differentiable on the interval $(a ; b)$ and $\lim_{x \rightarrow a+0} y^{\prime }(x) = \infty$, then $\lim_{x \rightarrow a+0} y(x) = \infty$;

if the function $y(x)$ is differentiable on the interval $(a ; b)$ and $\lim_{x \rightarrow a+0} y(x) = \infty$, then $\lim_{x \rightarrow a+0} y^{\prime }(x) = \infty$;

if the function $y(x)$ is differentiable on the interval $(a ;+\infty )$ and $\lim_{x \rightarrow +\infty } y(x)$ exists, then $\lim_{x \rightarrow +\infty } y^{\prime }(x)$ also exists;

if the function $y(x)$ is differentiable on the interval $(a ;+\infty )$ and $\lim_{x \rightarrow +\infty } y^{\prime }(x)$ exists, does a finite or infinite $\lim_{x \rightarrow +\infty } y(x)$ exist?

$y(x) = \begin{cases} 1-\cos x, & \text{ if } \quad x < 0 \\ \ln (1+x)-x, & \text{ if } \quad x \geqslant 0 \end{cases}$

$y(x) = \begin{cases} 2 x \cos x, & \text{ if } x < 0, \\ \sin 2 x, & \text{ if } x \geqslant 0 ;\end{cases}$

$y(x) = \begin{cases} \operatorname {sh} x-x, & \text{ if } x < 0 \\ x-\sin x, & \text{ if } x \geqslant 0\end{cases}$

$y(x) = \begin{cases} \operatorname {sh} x, & \text{ if } x < 0 \\ \sin x \operatorname {ch} x, & \text{ if } x \geqslant 0 \end{cases}$

$y(x) = \begin{cases} x^{10}, & \text{ if } x \text{ — rational number, } \\ -x^{10}, & \text{ if } x \text{ — irrational number; } \end{cases}$

$y(x) = \begin{cases} x^{100} \sin (1 / x), & \text{ if } x \neq 0 \\ 0, & \text{ if } x = 0 \end{cases}$

$y(x) = \begin{cases} e^{-1 / x^{2}}, & \text{ if } x \neq 0, \\ 0, & \text{ if } x = 0. \end{cases}$

$f$ is continuous if and only if $a > 0$.

$f^{\prime }(0)$ exists if and only if $a > 1$.

$f^{\prime }$ is bounded if and only if $a \geq 1+c$.

$f^{\prime }$ is continuous if and only if $a > 1+c$.

$f^{\prime \prime }(0)$ exists if and only if $a > 2+c$.

$f^{\prime \prime }$ is bounded if and only if $a \geq 2+2 c$.

$f^{\prime \prime }$ is continuous if and only if $a > 2+2 c$.