高数强化NO7|导数与微分|分段函数|具体函数|抽象函数|可导的充要条件|微分|偏导

对导数定义式的直接考查

分段函数的导数

$$\{\begin{array}{l}\boxed{显式分段函数例（31,32）}\{\begin{array}{l}\text{分段点用定义}\\ \text{非分段点用求导法则}\end{array}\\ \\ \boxed{隐式分段函数（含绝对值、最值等）}\{\begin{array}{l}①\text{转化为显式分段函数}\\ ②\text{绝对值（例33,34,35,36）}\\ （符号、绝对值、最大值、最小值、取整等）\end{array}\end{array}$$

$$\begin{array}{rl}& \text{解：}x=0\text{时，}g(0)={\int }_0^{1}f(0\cdot t)dt={\int }_0^{1}f(0)dt=f(0)=\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\frac{f(x)}{x}\cdot x=1\cdot 0=0；\\ & x\ne 0\text{时，}g(x)={\int }_0^{1}f(xt)dt\stackrel{u=xt}{=}{\int }_0^{x}f(u)d\frac{u}{x}⟹\frac{1}{x}{\int }_0^{x}f(u)du,\\ & \text{故}g(x)=\{\begin{array}{ll}\frac{1}{x}{\int }_0^{x}f(u)du& x\ne 0\\ 0& x=0\end{array}\\ & \\ & g^{\prime }(0)=\underset{x\to 0}{\lim }\frac{g(x)-g(0)}{x-0}=\underset{x\to 0}{\lim }\frac{\frac{1}{x}{\int }_0^{x}f(u)du}{x}\stackrel{\text{洛必达}}{=}\underset{x\to 0}{\lim }\frac{f(x)}{2x}=\frac{1}{2}；\\ & \\ & g^{\prime }(x)=\{\begin{array}{ll}\frac{xf(x)-{\int }_0^{x}f(u)du}{x^2}& x\ne 0\\ \frac{1}{2}& x=0\end{array}\\ & \\ & \text{验证}{g}^{\prime }(x)\text{在}x=0\text{处连续：}\\ & \underset{x\to 0}{\lim }{g}^{\prime }(x)=\underset{x\to 0}{\lim }\frac{xf(x)-{\int }_0^{x}f(u)du}{x^2}=\underset{x\to 0}{\lim }\frac{f(x)}{x}-\underset{x\to 0}{\lim }\frac{{\int }_0^{x}f(u)du}{x^2}=1-\frac{1}{2}=\frac{1}{2}=g^{\prime }(0),\\ & \text{故}{g}^{\prime }(x)\text{在}x=0\text{处连续.}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\\ & 1、\text{分段函数在分段点处导数的计算或是可导性的判断，一般直接用导数的定义。}\\ & \\ & 2、\text{函数}f(x)\text{可导，则函数}f(x)\text{连续。但是函数}f(x)\text{连续得不到}f(x)\text{可导。}\end{array}$$

$$\text{解1：}A、B\text{无可能，}C\text{概率小，}D\text{概率大，选}D.$$
$$\text{解2：}画图像，x\le 0是x，x>0时，当n=1，在\frac{1}{2}到1时1，在\frac{1}{3}到\frac{1}{2}是\frac{1}{2}.$$
$$线段一直往下，也一直变短，高度宽度都趋近于0，无限接近与原点，所以x\to 0时，图像光滑可导，选D$$
$$\begin{array}{rl}& \text{解3：}\underset{x\to 0^{-}}{\lim }\frac{f(x)-f(0)}{x-0}=1\\ & \\ & \underset{x\to 0^{+}}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{n}}{x}\\ & 1=\frac{\frac{1}{n}}{\frac{1}{n}}\le \frac{\frac{1}{n}}{x}<\frac{\frac{1}{n}}{\frac{1}{n+1}}\to 1\\ & =1\end{array}$$

$$解1：画图像，如果f(a)>0或f(a)<0，|f(x)|均可导，若f(a)=0，且f^{\prime }(a)=0，|f(x)|可导$$
$$f(a)只有斜穿x轴，|f(x)|小于0的部分翻上去之后，两边导数不一样，则|f(x)|不可导$$
$$选B$$
$$\begin{array}{rl}& \text{若}f(a)=0,f^{\prime }(a)\ne 0,\text{则：}\\ & \\ & \underset{x\to a^{+}}{\lim }\frac{|f(x)|-|f(a)|}{x-a}=\underset{x\to a^{+}}{\lim }\frac{|f(x)|}{|x-a|}=\underset{x\to a^{+}}{\lim }\left|\frac{f(x)}{x-a}\right|=\left|\underset{x\to a^{+}}{\lim }\frac{f(x)}{x-a}\right|=|f^{\prime }(a)|\\ & \\ & \underset{x\to a^{-}}{\lim }\frac{|f(x)|-|f(a)|}{x-a}=\underset{x\to a^{-}}{\lim }\frac{|f(x)|}{-|x-a|}=-\underset{x\to a^{-}}{\lim }\left|\frac{f(x)}{x-a}\right|=-\left|\underset{x\to a^{-}}{\lim }\frac{f(x)}{x-a}\right|=-|f^{\prime }(a)|\\ & \\ & \text{若}{f}^{\prime }(a)\ne 0,\text{则}|f(x)|\text{在}x=a\text{处不可导}\end{array}$$

$$\begin{array}{rl}& \text{解：}\\ & \text{若}f(x)\text{在}x=1\text{处可导，则}{f}_{+}^{\prime }(1)=f_{-}^{\prime }(1).\\ & \\ & f_{+}^{\prime }(1)=\underset{x\to 1^{+}}{\lim }\frac{f(x)-f(1)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{(x^3-1)\phi (x)}{x-1}=\underset{x\to 1^{+}}{\lim }(x^2+x+1)\phi (x)=3\phi (1)\\ & \\ & f_{-}^{\prime }(1)=\underset{x\to 1^{-}}{\lim }\frac{f(x)-f(1)}{x-1}=\underset{x\to 1^{-}}{\lim }\frac{(1-x^3)\phi (x)}{x-1}=-\underset{x\to 1^{-}}{\lim }(x^2+x+1)\phi (x)=-3\phi (1)\\ & \\ & \text{由}{f}_{+}^{\prime }(1)=f_{-}^{\prime }(1)⟹3\phi (1)=-3\phi (1)⟹\phi (1)=0,\text{故}\phi (1)=0\text{是}f(x)\text{在}x=1\text{处可导的必要条件.}\\ & \\ & \text{若}\phi (1)=0,\text{则}{f}_{+}^{\prime }(1)=3\phi (1)=0,f_{-}^{\prime }(1)=-3\phi (1)=0,\text{故}\phi (1)=0\text{是}f(x)\text{在}x=1\text{处可导的充分条件.}\\ & \\ & \text{综上，}\phi (1)=0\text{是}f(x)\text{在}x=1\text{处可导的充要条件，选}(A)\end{array}$$
$$\begin{array}{rl}& A⟹B,\text{则：}\\ & A\text{是}B\text{的充分条件，}B\text{是}A\text{的必要条件；}\\ & B\text{的充分条件是}A,A\text{的必要条件是}B.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\\ & f(x)\text{在}x=x_0\text{处连续，定义}g(x)=|x-x_0|f(x)，则：\\ & g(x)\text{在}x=x_0\text{处可导的充要条件是}f(x_0)=0.\end{array}$$
$$\begin{array}{rl}& \boxed{结论应用：下列函数的不可导点个数}\\ & ①f(x)=x^2-3x+20个不可导点\\ & ②f(x)=|x^2-3x+2|2个不可导点：x=2,x=1\\ & ③f(x)=(x-1)|x^2-3x+2|1个不可导点：x=2\\ & ④f(x)=x(x-1)|x^2-3x+2|1个不可导点：x=2\end{array}$$

$$\begin{array}{rl}& \text{解：}g(x)=x^2|x|=\{\begin{array}{ll}{x}^3& x>0\\ -x^3& x\le 0\end{array}\\ & \\ & g_{+}^{\prime }(0)=\underset{x\to 0^{+}}{\lim }\frac{x^3-0}{x}=0,g_{-}^{\prime }(0)=\underset{x\to 0^{-}}{\lim }\frac{-x^3-0}{x}=0\\ & \\ & g^{\prime }(x)=\{\begin{array}{ll}3x^2& x>0\\ -3x^2& x\le 0\end{array}\\ & \\ & g_{+}^{\prime \prime }(0)=\underset{x\to 0^{+}}{\lim }\frac{3x^2-0}{x}=0,g_{-}^{\prime \prime }(0)=\underset{x\to 0^{-}}{\lim }\frac{-3x^2-0}{x}=0\\ & \\ & g^{\prime \prime }(x)=\{\begin{array}{ll}6x& x>0\\ -6x& x\le 0\end{array}\\ & \\ & g_{+}^{\prime \prime \prime }(0)=\underset{x\to 0^{+}}{\lim }\frac{6x-0}{x}=6,g_{-}^{\prime \prime \prime }(0)=\underset{x\to 0^{-}}{\lim }\frac{-6x-0}{x}=-6\\ & \\ & \text{故}{g}^{\prime \prime \prime }(0)\text{不存在.}\\ & 选C\end{array}$$

$$\begin{array}{rl}& \text{解：}\\ & (A)\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\lim }\frac{|x|\sin |x|}{x}=\underset{x\to 0}{\lim }\frac{|x|\cdot |x|}{x}=\underset{x\to 0}{\lim }\frac{x^2}{x}=0,\text{在}x=0\text{处可导}\\ & \\ & (B)\underset{x\to 0}{\lim }\frac{|x|\sin \sqrt{|x|}}{x}=\underset{x\to 0}{\lim }\frac{|x|\sqrt{|x|}}{x}=\underset{x\to 0}{\lim }\frac{|x{|}^{\frac{3}{2}}}{x}=0,\text{在}x=0\text{处可导}\\ & \\ & (C)\underset{x\to 0}{\lim }\frac{\cos |x|-1}{x}=\underset{x\to 0}{\lim }\frac{-\frac{|x{|}^2}{2}}{x}=0,\text{在}x=0\text{处可导}\\ & \\ & (D)\underset{x\to 0}{\lim }\frac{\cos \sqrt{|x|}-1}{x}=\underset{x\to 0}{\lim }\frac{-\frac{(\sqrt{|x|}{)}^2}{2}}{x}=-\frac{1}{2}\underset{x\to 0}{\lim }\frac{|x|}{x},\text{该极限不存在，故在}x=0\text{处不可导}\\ & 选D\end{array}$$
$$\begin{array}{rl}& 小结：\\ & \text{已知}f(0)=0,f(x)\text{在}x=0\text{处可导，则}{f}^{\prime }(0)=\underset{x\to 0}{\lim }\frac{f(x)}{x}.\\ & \\ & ①\text{若}f(x)\text{由}x\text{所构成（无绝对值）：}\\ & \text{当阶数}\ge 1\text{时，}f(x)\text{在}x=0\text{处可导；当阶数}<1\text{时，不可导.}\\ & \\ & ②\text{若}f(x)\text{由}x\text{的绝对值构成：}\\ & \text{当}f(|x|)\text{的阶数}>1\text{时，}f(x)\text{在}x=0\text{处可导；当阶数}\le 1\text{时，不可导.}\end{array}$$

具体函数在指定点的导数

$$\begin{array}{rl}& \text{解：}\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\lim }\frac{x(x+1)(x+2)\cdots (x+n)}{x}=n!\end{array}$$
$$\begin{array}{rl}& \text{改题：}f(x)=\arctan \frac{e^{\sin (x-1)}-1}{\sqrt{1+x\sqrt{x^2-x+1+\sin \pi x}}},\text{求}{f}^{\prime }(1)\\ & \\ & f^{\prime }(1)=\underset{x\to 1}{\lim }\frac{\arctan \frac{e^{\sin (x-1)}-1}{\sqrt{1+x\sqrt{x^2-x+1+\sin \pi x}}}}{x-1}\\ & =\underset{x\to 1}{\lim }\frac{\frac{x-1}{\sqrt{2}}}{x-1}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\\ & \\ & \text{【答案】}\frac{\sqrt{2}}{2}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\\ & \text{如果函数}f(x)\text{可导，且满足}f(a)=0，则计算f^{\prime }(a)\text{时可以考虑使用导数的定义。}\end{array}$$

$$\begin{array}{rl}& \text{解：}f(x)=x^{\frac{2}{3}}\sin x\\ & \\ & f^{\prime }(0)=\underset{x\to 0}{\lim }\frac{\sqrt[3]{x^2}\sin x}{x}=0\\ & \\ & x\ne 0\text{时，}\\ & f^{\prime }(x)=\frac{2}{3}{x}^{-\frac{1}{3}}\sin x+x^{\frac{2}{3}}\cos x\\ & =\frac{2\sin x}{3\sqrt[3]{x}}+x^{\frac{2}{3}}\cos x\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{具体函数求导需要用到导数定义的两种常见情形：}\\ & (1)f(x)\text{表达式比较复杂，按照求导法则计算较为麻烦；}\\ & (2)\text{求导公式不成立的点。}\end{array}$$
$$\begin{array}{rl}& \text{小结：}\\ & ①\text{简单的、数值型函数求导一律使用求导法则，但若求导后出现定义域缩小，则对于缩小的点，用导数定义；}\\ & ②\text{分段点处用定义；}\\ & ③f(x)\text{表达式较复杂且}f(a)=0\text{时，也常用导数定义.}\end{array}$$

抽象函数的可导性

$$\begin{array}{rl}& (A)f(0)=\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\frac{f(x)}{x}\cdot x=0\\ & \\ & (B)f(0)=\frac{1}{2}\underset{x\to 0}{\lim }\left[f(x)+f(-x)\right]=\frac{1}{2}\underset{x\to 0}{\lim }\frac{f(x)+f(-x)}{x}\cdot x=0\\ & \\ & \Rightarrow f(0)=0\\ & (C)f^{\prime }(0)=\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\lim }\frac{f(x)}{x}\end{array}$$
$$\begin{array}{rl}& (D)\underset{x\to 0}{\lim }\frac{f(x)-f(0)-\left(f(-x)-f(0)\right)}{x}\stackrel{①}{=}\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x}+\underset{x\to 0}{\lim }\frac{f(-x)-f(0)}{-x}=f^{\prime }(0)+f^{\prime }(0)\\ & =[2f^{\prime }(0)]\\ & \\ & ①:\text{能拆分吗？}⟸\text{不一定存在啊，不能拆}\\ & 选D\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\\ & (1)\text{极限收敛性讨论中常用的两个法则：}\\ & ①\text{如果极限}\underset{x\to \square }{\lim }\frac{f(x)}{g(x)}\text{存在，且}\underset{x\to \square }{\lim }g(x)=0，则有\underset{x\to \square }{\lim }f(x)=0；\\ & ②\text{如果极限}\underset{x\to \square }{\lim }\frac{f(x)}{g(x)}\text{存在且不为零，且}\underset{x\to \square }{\lim }f(x)=0，则有\underset{x\to \square }{\lim }g(x)=0。\\ & \\ & (2)\text{本题}(C)\text{选项的结论可以记住：如果}f(x)\text{连续，且有}\underset{x\to a}{\lim }\frac{f(x)-A}{x-a}=b，\\ & \text{则可以得到}f(x)\text{在}x=a\text{处可导且}f(a)=A,f^{\prime }(a)=b。\end{array}$$

$$\begin{array}{rl}& \text{与}f(0)=?\text{无关，可以对}f(0)\text{任意赋值，使得}\\ & f(x)\text{在}x=0\text{处不连续，自然不可导，排除}A、B.\\ & \\ & f(x)=\{\begin{array}{ll}\square & x\ne 0\\ 1,2,3,\cdots & x=0\end{array}\\ & \\ & \text{若}D\text{对，则}C\text{对，故选}(C)\end{array}$$
$$\begin{array}{rl}& f(x)\text{在}x=0\text{处可导，则}f(x)\text{在}x=0\text{处连续，}\\ & f(0)=\underset{x\to 0}{\lim }f(x)=0,\underset{x\to 0}{\lim }\frac{f(x)}{x}\text{存在.}\\ & \\ & \underset{x\to 0}{\lim }\frac{f(x)}{x}\cdot \frac{x}{\sqrt{x+1}}=0\end{array}$$

导数与极限

$$\begin{array}{rl}& \text{解：}\underset{x\to 0}{\lim }\frac{x^{2}f(x)-2f(x^3)}{x^3}=\underset{x\to 0}{\lim }\frac{x^{2}f(x)}{x^3}-\underset{x\to 0}{\lim }\frac{2f(x^3)}{x^3}\\ & =\underset{x\to 0}{\lim }\frac{f(x)}{x}-2\underset{x\to 0}{\lim }\frac{f(x^3)-f(0)}{x^3-0}\\ & =f^{\prime }(0)-2f^{\prime }(0)=-f^{\prime }(0)\\ & \text{选}(B).\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{考试中已知函数可导，求极限时往往用的是导数定义的广义化形式；}\\ & f^{\prime }(x_0)=\underset{\Delta x\to 0}{\lim }\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}=\underset{\square \to 0}{\lim }\frac{f(x_0+\square )-f(x_0)}{\square }\end{array}$$

可导的充要条件

$$\begin{array}{rl}& (A)\underset{h\to 0}{\lim }\frac{f(1-\cosh )}{h^2}=\underset{h\to 0}{\lim }\frac{f(1-\cosh )-f(0)}{1-\cosh -0}\cdot \frac{1-\cosh }{h^2}\\ & =\frac{1}{2}\underset{h\to 0}{\lim }\frac{f(1-\cosh )-f(0)}{1-\cosh -0}\stackrel{t=1-\cosh }{=}\frac{1}{2}\underset{t\to 0^{+}}{\lim }\frac{f(t)-f(0)}{t-0}\\ & =f_{+}^{\prime }(0).\\ & \\ & (B)\underset{h\to 0}{\lim }\frac{f(1-e^h)-f(0)}{1-e^h}\cdot \frac{1-e^h}{h}=-f^{\prime }(0)\\ & \\ & (C)不同阶\\ & \\ & (D)不能拆，不能确定导数存不存在\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}f(x)\text{在点}{x}_0\text{可导的充要条件是}\underset{\square \to 0}{\lim }\frac{f(x_0+\square )-f(x_0)}{\Delta }\text{存在，其中}\square \text{与}\Delta \text{为同阶无穷小。}\\ & \text{总结三点：①分子一动一静；②分子分母同阶；③分母可正可负。}\end{array}$$

$$因为函数在x=0处连续，所以f(0)=0，因为分母是h^2，所以是0^{+}，选C$$
$$\begin{array}{rl}& \text{改题：}\\ & \underset{n\to \infty }{\lim }(n+1)f\left(\frac{n}{n+1}\right)=1\end{array}$$
$$\begin{array}{rl}& \underset{n\to \infty }{\lim }\frac{f\left(\frac{n}{n+1}\right)}{\frac{1}{n+1}}\text{存在}⟹f(1)=0\\ & \\ & \underset{n\to \infty }{\lim }\frac{f\left(\frac{n}{n+1}\right)-f(1)}{\frac{n}{n+1}-1}\cdot \frac{\frac{n}{n+1}-1}{\frac{1}{n+1}}=-f_{-}^{\prime }(1)\\ & \text{该极限存在}⟹f_{-}^{\prime }(1)\text{存在}\\ & \\ & n\to \infty \text{时，}\frac{n}{n+1}\to 1^{-},\frac{n}{n+1}<1\end{array}$$
$$\begin{array}{rl}& 改：\\ & \underset{x\to 1}{\lim }\frac{f\left((x-1)\sin \pi x\right)}{(x-1)\sin \pi x}\text{存在}⟹f^{\prime }(0)\text{存在}\\ & \\ & \stackrel{t=(x-1)\sin \pi x}{=}\underset{t\to 0^{-}}{\lim }\frac{f(t)}{t}=f_{-}^{\prime }(0)\text{存在}\\ & \\ & x\to 1^{+}\text{时：}x-1\to 0^{+},\sin \pi x\to 0^{-}⟹(x-1)\sin \pi x\to 0^{-}\\ & x\to 1^{-}\text{时：}x-1\to 0^{-},\sin \pi x\to 0^{+}⟹(x-1)\sin \pi x\to 0^{-}\end{array}$$

微分

$$\{\begin{array}{l}\text{代数意义：}\Delta y=A\Delta x+o(\Delta x)\\ \text{几何意义：}\Delta y\text{：曲线的增量，}dy\text{：切线的增量.}\end{array}$$

$$\begin{array}{rl}& A\Delta x=y^{\prime }(x_0)\Delta x\\ & \\ & \text{解：}{y}^{\prime }={\left(f(x^2)\right)}^{\prime }=f^{\prime }(x^2)\cdot 2x,\\ & y^{\prime }(-1)=f^{\prime }\left((-1{)}^2\right)\cdot 2\cdot (-1)=-2f^{\prime }(1)\\ & \\ & -2f^{\prime }(1)\cdot (-0.1)=0.1⟹f^{\prime }(1)=0.5\\ & 选D\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{关于}dy\text{与}\Delta y：\Delta y\text{是曲线垂直方向增量，}dy\text{是曲线沿切线垂直方向的增量。}\end{array}$$

多元函数的极限、连续、偏导与微分

$$\begin{array}{rl}& \text{解：}①{\frac{\partial f}{\partial x}|}_{(0,0)}=\underset{x\to 0}{\lim }\frac{f(x,0)-f(0,0)}{x-0}=\underset{x\to 0}{\lim }\frac{x-0}{x-0}=1\\ & \\ & ③\underset{(x,y)\to (0,0)}{\lim }f(x,y)=0.(\text{无论选取}xy,x,y\text{中的任意一个，极限都是}0)✓\\ & \\ & (x,y)\to (0,0)\text{时：}\\ & ①沿y=x：f(x,y)=xy；②沿x：f(x,y)=x；③沿y：f(x,y)=y\end{array}$$
$$\begin{array}{rl}& ④\text{（}x\text{无限接近但不等于}0\text{）、（}y\text{无限接近但不等于}0\text{）}\\ & \text{故}\underset{y\to 0}{\lim }\underset{x\to 0}{\lim }f(x,y)=\underset{y\to 0}{\lim }\underset{x\to 0}{\lim }xy=0\end{array}$$
$$\begin{array}{rl}& ②\text{先求}\frac{\partial f}{\partial x}\text{，再求}{\frac{{\partial }^{2}f}{\partial x\partial y}|}_{(0,0)}\\ & \frac{\partial f}{\partial x}=\underset{\Delta x\to 0}{\lim }\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}\end{array}$$
$$\begin{array}{rl}& ①\text{若}y=0,\\ & \underset{\Delta x\to 0}{\lim }\frac{f(x+\Delta x,0)-f(x,0)}{\Delta x}=\underset{\Delta x\to 0}{\lim }\frac{x+\Delta x-x}{\Delta x}=1\\ & \\ & ②\text{若}y\ne 0,x\ne 0,\\ & \underset{\Delta x\to 0}{\lim }\frac{(x+\Delta x)\cdot y-xy}{\Delta x}=y\\ & \\ & ③\text{若}y\ne 0,x=0,\\ & \underset{\Delta x\to 0}{\lim }\frac{f(\Delta x,y)-y}{\Delta x}=\underset{\Delta x\to 0}{\lim }\frac{\Delta x\cdot y-y}{\Delta x}=\infty ,\\ & \text{故}{\frac{{\partial }^{2}f}{\partial x\partial y}|}_{(0,0)}\text{不存在}.\end{array}$$
$$\begin{array}{rl}& \text{考试：}②{\frac{{\partial }^{2}f}{\partial x\partial y}|}_{(0,0)}={\frac{\partial \left(\frac{\partial f}{\partial x}\right)}{\partial y}|}_{(0,0)}=\underset{y\to 0}{\lim }\frac{f_x^{\prime }(0,y)-f_x^{\prime }(0,0)}{y-0}(y\ne 0)\\ & \\ & f_x^{\prime }(0,y)=\underset{x\to 0}{\lim }\frac{f(x,y)-f(0,y)}{x-0}(x\ne 0,y\ne 0)\\ & =\underset{x\to 0}{\lim }\frac{xy-y}{x}=\infty \text{，不存在}.\end{array}$$
$$选B$$
$$\begin{array}{rl}& \text{改题：}\\ & \text{关于函数}f(x,y)=\{\begin{array}{ll}xy,& xy\ne 0,\\ x+1,& y=0,\\ y+2,& x=0,\end{array}\text{给出以下结论}\\ & \\ & ①\underset{x\to 0}{\lim }f(x,y)\text{存在且为}0.\\ & ②\underset{x\to 0}{\lim }f(x,y)\text{存在且为}1.\\ & ③\underset{x\to 0}{\lim }f(x,y)\text{不存在}.✓\end{array}$$
$$\begin{array}{rl}& x\to 0\text{表示}x\text{无限接近但不等于}0,\\ & \text{但对}y\text{无要求：}\\ & y=0\text{时，取}x+1；\\ & y\ne 0\text{时，取}xy.\\ & 极限不相等，所以不存在，选3\end{array}$$
$$\begin{array}{rl}& \text{再判断}\\ & ①\underset{y\to 0}{\lim }f(x,y)=2.\\ & ②\underset{y\to 0}{\lim }f(x,y)=0.\\ & ③\underset{y\to 0}{\lim }f(x,y)\text{不存在.}✓\\ & \\ & \underset{x\to 0}{\lim }\underset{y\to 0}{\lim }f(x,y)=0✓\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{二元函数连续、可偏导、全微分三者间的关系：}\\ & \\ & \text{连续}\nleftrightarrow \text{（偏导数存在）}\\ & \\ & \text{可微}\leftarrow \text{偏导函数连续}\{\begin{array}{l}{\lim }_{(x,y)\to (x_0,y_0)}{f}_x^{\prime }(x,y)=f_x^{\prime }(x_0,y_0)\\ {\lim }_{(x,y)\to (x_0,y_0)}{f}_y^{\prime }(x,y)=f_y^{\prime }(x_0,y_0)\end{array}\\ & \\ & \text{定义：}\underset{(\Delta x,\Delta y)\to (0,0)}{\lim }\frac{\Delta z-A\Delta x-B\Delta y}{\sqrt{(\Delta x{)}^2+(\Delta y{)}^2}}=0\\ & 连续\nleftrightarrow 可微\\ & 可微\nleftrightarrow （偏导数存在）\\ & 可微\to \leftarrow 定义\end{array}$$