微积分作业代写calclulus代考| Differentiability of f : Rn −→ R

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微积分作业代写calclulus代考|Example 2.9:

Consider the function $f: \mathbb{R}^{2} \mapsto \mathbb{R}, f(x, y)=x \arctan (y / x), f(0, y)=0$. We wish to discuss the continuity of $f$, and the existence and continuity of $f_{x}$ and $f_{y}$ at points on the $x$-axis.

We have $\lim {x \rightarrow 0} f(x, y)=\lim {x \rightarrow 0} x \arctan (y / x)=0=f(0, y)$ for all $y$, since $|\arctan (y / x)|<\pi / 2$ The function is continuous for all points on the $x$-axis. We have, for $y \neq 0, f_{x}=\arctan (y / x)-\frac{x y}{x^{2}+y^{2}}$. Then for $y>0, \lim {x \rightarrow 0^{+}} f{x}=\frac{\pi}{2}-0=\frac{\pi}{2}, \quad \lim {x \rightarrow 0^{-}} f{x}=-\frac{\pi}{2}+0=-\frac{\pi}{2}$; and for $y<0, \lim {x \rightarrow 0^{+}} f{x}=-\frac{\pi}{2}+0=-\frac{\pi}{2}, \quad \lim {x \rightarrow 0^{-}} f{x}=\frac{\pi}{2}-0=\frac{\pi}{2}$.
Thus $f_{x}$ is not continuous along $x=0$ for $y \neq 0$. Also,
$$f_{y}=\frac{x^{2}}{x^{2}+y^{2}}, \quad \lim {x \rightarrow 0^{+}} f{y}=\lim {x \rightarrow 0^{-}} f{y}=0 .$$
If we define $f_{\nu}(0,0)=0$, then $f_{\nu}$ exists and is continuous on $x=0$.

微积分作业代写calclulus代考|Mastery Check 2.8:

Consider the function $f: \mathbb{R}^{2} \longrightarrow \mathbb{R}$ defined by $z=f(x, y)=\frac{x y}{\sqrt{x^{2}+y^{2}}}$ if $x^{2}+y^{2} \neq 0, f(0,0)=0$,
whose graph appears at the end of this exercise in Figure 2.15. We wish to investigate the behaviour of $f$ near $(0,0)$.
(1) Find the two partial derivatives for $(x, y) \neq(0,0)$.
(2) Show using Definition $2.3$ that both partial derivatives are zero at $(0,0)$. Thus, the tangent plane at $(0,0)$, if it exists, must coincide with the plane $z=0$.

2.D Differentiability
73
(3) But the partial derivatives are not continuous. Show that these limits are not all the same:
$$\lim {x \rightarrow 0}\left(\lim {y \rightarrow 0} \frac{\partial f}{\partial x}\right), \quad \lim {y \rightarrow 0^{-}}\left(\lim {x \rightarrow 0} \frac{\partial f}{\partial x}\right), \quad \lim {y \rightarrow 0^{+}}\left(\lim {x \rightarrow 0} \frac{\partial f}{\partial x}\right) .$$
(A similar result holds for the other derivative.)
So, we do not expect the tangent plane to exist.
(4) Now recall the properties of a tangent plane as outlined in Definition 2.4. See if you can construct the expression $\Delta z=f(\boldsymbol{x})-\phi\left(\boldsymbol{x}-\boldsymbol{x}{0}\right)=f\left(\boldsymbol{x}{0}\right)+|\boldsymbol{\Delta} \boldsymbol{x}| \rho(\boldsymbol{\Delta} \boldsymbol{x}) \quad$ for the special case that $\boldsymbol{x}$ lies on the line $y=x$, at distance
$|\Delta \boldsymbol{x}|=\sqrt{\Delta x^{2}+\Delta y^{2}}$ from $\boldsymbol{x}_{0}=(0,0)$, with $\Delta y=\Delta x .$
That is, find $\rho(\Delta \boldsymbol{x})$.
Use this result to decide whether a tangent plane exists.

微积分作业代写calclulus代考|Example 2.9:

Consider the function f:R2↦R,f(x,y)=xarctan⁡(y/x),f(0,y)=0. We wish to discuss the continuity of f, and the existence and continuity of fx and fy at points on the x-axis.

f_{y}=\frac{x^{2}}{x^{2}+y^{2}}, \quad \lim {x \rightarrow 0^{+}} f {y}=\lim {x \rightarrow 0^{-}} f {y}=0 。
$$如果我们定义Fν(0,0)=0， 然后Fν存在并且持续X=0. 微积分作业代写calclulus代考|Mastery Check 2.8: 考虑函数F:R2⟶R被定义为和=F(X,和)=X和X2+和2如果X2+和2≠0,F(0,0)=0， 其图表出现在图 2.15 中本练习的末尾。我们希望调查的行为F靠近(0,0). (1) 求两个偏导数(X,和)≠(0,0). (2) 使用定义显示2.3两个偏导数都为零(0,0). 因此，切平面在(0,0)，如果存在，必须与平面重合和=0. 2.D Differentiability 73 (3) But the partial derivatives are not continuous. Show that these limits are not all the same:$$
\lim {x \rightarrow 0}\left(\lim {y \rightarrow 0} \frac{\partial f}{\partial x}\right), \quad \lim {y \rightarrow 0^{-}}\left(\lim {x \rightarrow 0} \frac{\partial f}{\partial x}\right), \quad \lim {y \rightarrow 0^{+}}\left(\lim {x \rightarrow 0} \frac{\partial f}{\partial x}\right) .

(A similar result holds for the other derivative.)
So, we do not expect the tangent plane to exist.
(4) 现在回想定义 2.4 中概述的切平面的性质。看看你是否可以构造表达式 $\Delta z=f(\boldsymbol{x})-\phi\left(\boldsymbol{x}-\boldsymbol{x} {0}\right)=f\left(\boldsymbol {x} {0}\right)+|\boldsymbol{\Delta} \boldsymbol{x}| \rho(\boldsymbol{\Delta} \boldsymbol{x}) \quadF○r吨H和sp和C一世一种一世C一种s和吨H一种吨\boldsymbol{x}一世一世和s○n吨H和一世一世n和y=x,一种吨d一世s吨一种nC和|\Delta \boldsymbol{x}|=\sqrt{\Delta x^{2}+\Delta y^{2}}Fr○米\boldsymbol{x}_{0}=(0,0),在一世吨H\Delta y=\Delta x 。吨H一种吨一世s,F一世nd\rho(\Delta \boldsymbol{x})$。