Differentiability and the total differential

Show that $f$ is continuous at $(0,0)$.

Calculate all the directional derivatives of $f$ at $(0,0)$.

Show that $f$ is not differentiable at $(0,0)$.

$(1+x)^{m}(1+y)^{n}$;

$\ln (1+x) \cdot \ln (1+y)$;

$\arctan \dfrac {x+y}{1+xy}$.

$1.002 \cdot 2.003^{2} \cdot 3.004^{3}$;

$\dfrac {1.03^{2}}{\sqrt[3]{0.98}\sqrt[4]{1.05^{3}}}$;

$\sqrt{1.02^{3}+1.97^{3}}$;

$\sin 29^{\circ }\tan 46^{\circ }$;

$0.97^{1.05}$.

$(1 ; 1)$;

$(1 ; 0)$;

$(0 ; 0)$.

if the function has partial derivatives with respect to all variables at some point, then $f(x)$ is continuous at this point;

if the function has partial derivatives with respect to all variables at every point of the space $R^{n}$, then it is continuous in $R^{n}$;

if the function is differentiable at some point, then at this point the function has partial derivatives with respect to all variables;

if the function has partial derivatives with respect to all variables at some point, then the function is differentiable at this point;

if the function is differentiable at some point, then at this point the function has continuous partial derivatives with respect to all variables;

if the function has continuous partial derivatives at some point, then the function is differentiable at this point.

$f = 2 x^{4}-3 x^{2} y^{2}+x^{3} y$;

$f = \left(y^{3}+2 x^{2} y+3\right)^{4}$;

$f = \frac{y}{x}+\frac{x}{y}$

$f = \frac{x}{\sqrt{x^{2}+y^{2}}}$;

$f = 2^{-y / x}$;

$f = \ln \left(x+\sqrt{x^{2}+y^{2}}\right)$;

$f = \ln \sin \frac{x+1}{\sqrt{y}}$;

$f = \operatorname {arctg} \frac{x}{y}+\operatorname {arctg} \frac{y}{x}$;

$f = \operatorname {arctg} \frac{x+y}{x-y}$;

$f = (1+x y)^{y}$.

$f(x ; y) = (5 x+7 y-25) e^{-\left(x^{2}+x y+y^{2}\right)}$;

$f(x ; y ; z) = 2 y^{2}+z^{2}-x y^{2}-y z+4 x+1$.

$f = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$,

1. $(1 ; 1)$;

2. $(0 ; 1)$;

$f = \sqrt{x y+\frac{x}{y}},(2 ; 1)$;

$f = \frac{\cos (x-2 y)}{\cos (x+2 y)},\left(\frac{\pi }{4} ; \pi \right)$;

$f = 2^{\operatorname {tg}\left(\pi x /\left(x+3 y^{2}\right)\right)},(1 ; 1)$;

$f = \ln \frac{4 \sqrt{2-x}-x y}{1+\cos y},(1 ; 0)$;

$f = \arccos \sqrt{x^{2}-2 y},(1 ; 0,18)$;

$f = \operatorname {arctg} \frac{y}{1+x^{2}},(1 ;-1)$;

$f = e^{x-1} \operatorname {arctg} \frac{2 x+3 y}{1-6 x y}$, (1;1);

$f = \ln \arcsin \left(x+y^{3}\right)$, $\left(\frac{\sqrt{3}}{2} ; 0\right)$;

$f = \operatorname {arcctg} \ln \left(\sqrt{x}+y^{4}\right)$, $\left(e^{2} ; 0\right)$.

$f = \sqrt{x^{2}+y^{2}+z^{2}}$;

$f = e^{x y \sin z}$

$f = (x y)^{z}$

$f = x^{y / z}$.

$f = \sin \sum_{i = 1}^{n} x_{i}^{2}$;

$f = \ln \left(\left(1-\sqrt{\sum_{i = 1}^{n} x_{i}^{2}}\right) /\left(1+\sqrt{\sum_{i = 1}^{n} x_{i}^{2}}\right)\right)$.

$f = \frac{x}{x^{2}+y^{2}+z^{2}},(1 ; 0 ; 1)$;

$f = \operatorname {arctg} \frac{x y}{z^{2}},(3 ; 2 ; 1)$;

$f = \left(x y+\frac{x}{y}\right)^{z},(1 ; 1 ; 1)$;

$f = \ln \sum_{i = 1}^{n} x_{i},(1 ; 2 ; \ldots ; n)$.

$f = x(\sqrt[3]{1+\sqrt{\left|y\right|}}-1)$;

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

$f = (\sin x+\sqrt[3]{x y})^{2}$;

$f = \operatorname {ch} \sqrt[5]{x^{2} y}$;

$f = \sqrt[5]{x^{4}}(\cos \sqrt[5]{y}-1)$;

$f = \ln \left(2-|x|^{7 / 6}+|y|^{5 / 4}\right)$;

$f = \sqrt[3]{y^{2}} \operatorname {arctg} \sqrt{\left|x\right|}$;

$f = \begin{cases} x \sin (y / \sqrt{\left|x\right|}), & x \neq 0 \\ 0, & x = 0 \end{cases}$

$f = y^{3 / 5} \arcsin \sqrt{\left|x\right|}$;

$f = y+\operatorname {ch} \sqrt[3]{x^{2}+y^{2}}$;

$f = \sqrt{1-|x|^{2 / 3}|y|^{5 / 6}}$;

$f = 1+x y+\sin \sqrt[5]{x^{2} y^{4}}$.

$f = \sqrt{\left|x y\right|}$

$f = \sqrt{x^{2}+y^{2}}$;

$f = \sqrt[3]{x^{3}+y^{3}}$

$f = \sqrt[5]{\sin x(1-\cos x y)}$;

$f = \sin \left(\pi / 4+\sqrt[3]{x y^{2}}\right)$

$f = \ln \left(3+\sqrt[3]{x^{2} y}\right)$;

$f = \sin \left(3 x+\sqrt[3]{x^{3}-y^{3}}\right)$

$f = \ln \left(1+x+\sqrt[3]{x^{3}+27 y^{3}}\right)$;

$f = \operatorname {arctg}\left(2 x+\sqrt[3]{x^{3}-27 y^{3}}\right)$;

$f = \arcsin \left(y+\sqrt[3]{x^{3}+8 y^{3}}\right)$.

$f = y \sqrt[3]{1+\sqrt[3]{x^{2}}}$;

$f = \sqrt[3]{x^{3}+y^{4}}$;

$f = 2 y+x \cos \sqrt[3]{x y}$;

$f = y+\cos \sqrt[3]{x^{2}+y^{2}}$;

$f = \sqrt[3]{\sin^{4} x+\cos^{4} y}$;

$f = y+\ln \left(3+\sqrt[3]{x^{2} y}\right)$;

$f = \arcsin \left(x y+\sqrt[3]{x^{3}+y^{3}}\right)$;

$f = \operatorname {arctg}\left(x y+y+\sqrt[3]{x^{2} y}\right)$;

$f = \begin{cases} \left(y^{3}-x^{3}\right) /\left(x^{2}+2 y^{2}\right), & \text{ if } & x^{2}+y^{2} \neq 0, \\ 0, & \text{ if } & x^{2}+y^{2} = 0 \end{cases}$

$f = \begin{cases} (x+y) \operatorname {arctg}(x / y)^{2}, & \text{ if } y \neq 0, \\ \pi x / 2, & \text{ if } y = 0 \end{cases}$

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

$f = \sqrt[4]{\left|(x-1) y^{3}\right|}-x y$;

$f = x y+\sin \sqrt[5]{(x-1)^{2} y^{4}}$;

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

$f = x+2 y+e \sqrt[3]{(x-1)^{2} y}$.

Prove that the function

is differentiable, but not continuously differentiable, in $R^{2}$;

prove that the function $f = \begin{cases} \left(x^{2}+y^{2}\right)^{\alpha }, & \text{ if } x, y \text{-are rational numbers, } \\ 0, & \text{ if at least one of the numbers } x, y \end{cases}$ is irrational, for $\alpha > 1 / 2$ is differentiable only at the point $(0,0)$ and is not continuously differentiable at this point.

$f = |x+y|^{3 \alpha }+|y|^{4-\alpha } ;$

$f = \frac{|x|^{3 \alpha }+|y|^{7-\alpha }}{x^{2}+y^{2}}$;

$f = \frac{|x|^{\alpha }|y|^{5-2 \alpha }}{|x|+|y|}$;

$f = \frac{|x y|^{\alpha }}{1-\ln \left(x^{2}+y^{2}\right)}$.

$f = \frac{\left|x^{3} y\right|^{1 / 2}}{\left(x^{2}+x y+y^{2}\right)^{\alpha }}, \alpha = \frac{1}{2}, \alpha = \frac{1}{4}$;

$f = \frac{x^{3} y^{2}}{\left(x^{6}+y^{6}\right)^{\alpha }}, \alpha = \frac{1}{2}, \alpha = \frac{2}{3}$;

$f = \frac{\left(x^{2} y^{3}\right)^{3 / 5}}{\left(x^{2}-x y+y^{2}\right)^{\alpha }}, \alpha = 1, \alpha = \frac{1}{2}$;

$f = \frac{\left(x^{2} y\right)^{4 / 3}}{\left(x^{4}+y^{4}\right)^{\alpha }}, \alpha = \frac{1}{2}, \alpha = \frac{3}{4}$.

$f = x\left|y\right|+y\left|x\right|$

$f = (y-\left|x\right|)^{2}$;

$f = \left|x^{2}-y^{2}\right|$;

$f = \frac{1}{1+\left|x y\right|}$.

$f = \dfrac {x \ln (1+y)-y \ln (1+x)+(x y / 2)(y-x)}{\left(x^{2}+y^{2}\right)^{3 / 2}}$;

$f = \dfrac {x \operatorname {sh} y-y \operatorname {sh} x+(x y / 6)\left(x^{2}-y^{2}\right)}{\left(x^{2}+y^{2}\right)^{5 / 2}}$;

$f = \frac{x \operatorname {arctg} y-y \operatorname {arctg} x+(x y / 3)\left(y^{2}-x^{2}\right)}{\left(x^{2}+y^{2}\right)^{5 / 2}}$;

$f = \frac{x e^{y}-y e^{x}+y-x+(x y / 2)(x-y)}{\left(x^{2}+y^{2}\right)^{3 / 2}}$;

$f = \frac{\sqrt{1+x y}-e^{x y / 2}}{\left(x^{2}+y^{2}\right)^{3 / 2}}$;

$f = \frac{\left(x^{3} y+y^{5}\right)^{1 / 3}}{\left|x\right|+\left|y\right|}+2 y$.

$\varphi = f(u), u = x y+y^{2} / x$

$\varphi = f(u ; v), u = y /(x+y), v = x^{2}-y^{3}$;

$\varphi = f(u ; v), u = y^{2}, v = \operatorname {arctg}(y / x)$;

$\varphi = f(u ; v ; w), u = x^{2}+y^{2}+z^{2}, v = x+y+z, w = x y z$.

$u = y+\sqrt{1+|x|^{3}|y|^{3 / 2}}, \quad v = x+\sqrt{5-|x|^{3 / 2}|y|^{5 / 6}}$;

$u = \begin{cases} -\frac{x}{\ln^{2}\left|y\right|}, & y \neq 0, \\ 0, & y = 0, \end{cases} \quad v = \begin{cases} \sqrt{\left|x y\right|} \cos (1 / y), & y \neq 0, \\ 0, & y = 0. \end{cases}$

Prove that $D_{1} f$ and $D_{2} f$ are bounded functions in $\mathbb {R}^{2}$. (Hence $f$ is continuous).

Let $\mathbf{u}$ be any unit vector in $\mathbb {R}^{2}$. Show that the directional derivative $(D_{\mathbf{u}} f)(0,0)$ exists, and that its absolute value is at most 1.

Let $\gamma$ be a differentiable mapping of $\mathbb {R}^{1}$ into $\mathbb {R}^{2}$ (in other words, $\gamma$ is a differentiable curve in $\mathbb {R}^{2}$,, with $\gamma (0) = (0,0)$ and $\left|\gamma^{\prime }(0)\right| > 0$. Put $g(t) = f(\gamma (t))$ and prove that $g$ is differentiable for every $t \in \mathbb {R}^{1}$.

If $\gamma \in \mathscr {C}^{\prime }$, prove that $g \in \mathscr {C}^{\prime }$.

In spite of this, prove that $f$ is not differentiable at $(0,0)$.

Hint: Formula (40) fails.