Chapter 4 Differential Calculus of Functions of Several Variables
Section 1 Basic Concepts and Conclusions
Definition 1: (Function of Two Variables)
\[\begin{align} &z=f(x,y),(x,y)\in D\subset R^2\\ \end{align}\]
Example: \(\begin{align} &\text{Find the domain of } f(x,y)=\arcsin(2x)+\ln y+\frac{\sqrt{4x-y^2}}{\ln{(1-x^2-y^2)}}.\\ &\text{Solution:}\\ &-1\leq 2x\leq 1, \quad y > 0, \quad 1-x^2-y^2 > 0, \quad 1-x^2-y^2\neq 1, \quad 4x-y^2\geq 0\\ &D=\left\{(x,y)\,\middle|\,-\frac{1}{2}\leq x\leq\frac{1}{2}, \quad y > 0, \quad x^2+y^2 < 1, \quad x\geq\frac{1}{4}y^2\right\}\\ \end{align}\)
Definition 2: (Limit of a Function of Two Variables)
\[\begin{align} &\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=A \quad \text{or} \quad \lim_{x\rightarrow x_0, y\rightarrow y_0}f(x,y)=A\\ \end{align}\]Examples: \(\begin{align} &\text{1. Evaluate the limit:}\\ &\text{Solution Method 1:}\\ &\lim_{x\rightarrow 0,y\rightarrow 0}xy\frac{x^2-y^2}{x^2+y^2}\\ &\xlongequal[\text{Let } y=r\sin\theta]{\text{Let } x=r\cos\theta}\lim_{r\rightarrow0}{r\cos\theta \cdot r\sin\theta \cdot \frac{r^2\cos2\theta}{r^2}}=0\\ &\text{Solution Method 2:}\\ &0\leq\left|xy\frac{x^2-y^2}{x^2+y^2}\right|\leq{|xy|}\\ &\because \lim_{x\rightarrow 0,y\rightarrow 0}{0}=0 \quad \text{and} \quad \lim_{x\rightarrow 0,y\rightarrow 0}|xy|=0\\ &\text{By the Squeeze Theorem: } \Leftrightarrow\lim_{x\rightarrow 0,y\rightarrow 0}xy\frac{x^2-y^2}{x^2+y^2}=0\\ &\\ &\text{2. Verify that the following limits do not exist:}\\ &(1)\lim_{x\rightarrow 0,y\rightarrow 0}\frac{xy}{x^2+y^2}\\ &(2)\lim_{x\rightarrow 0,y\rightarrow 0}\frac{x^3+y^3}{x^2+y}\\ &\text{Solution:}\\ &(1) \text{ Let } y=kx:\\ &\lim_{x\rightarrow0,y=kx}\frac{x \cdot kx}{x^2+k^2x^2}=\frac{k}{1+k^2} \quad \text{(The limit depends on } k\text{, so it does not exist.)}\\ &(2) \text{ Let } y=-x^2+x^4:\\ &\lim_{x\rightarrow0,y=-x^2+x^4}\frac{x^3+(x^4-x^2)^3}{x^2+(-x^2+x^4)}=\lim_{x\rightarrow0}\left[\frac{1}{x}+\frac{(x^4-x^2)^3}{x^4}\right]=\infty \quad \text{(The limit does not exist.)}\\ \end{align}\) Notes: \(\begin{align} &\text{Function of a single variable: } \{(x,f(x))|x\in D\}\\ &\text{Function of several variables: } \{(x,y,f(x,y))|(x,y)\in D\} \end{align}\)
Definition 3: (Continuity of a Function of Two Variables)
\[\begin{align} &f(x,y) \text{ is continuous at point } P_0: \lim_{x\rightarrow x_0,y\rightarrow y_0}f(x,y)=f(x_0,y_0)\\ &\text{Note 1: } z=f(x,y) \text{ is continuous at point } P_0 \Leftrightarrow \Delta z=f(x,y)-f(x_0,y_0)\rightarrow 0 \text{ as } (x\rightarrow x_0,y\rightarrow y_0)\\ &\text{Note 2: "Elementary functions of two variables" are continuous everywhere within their domains. e.g., } \lim_{x\rightarrow 1,y\rightarrow2}\frac{x+y}{x-y}=-3\\ \end{align}\]Theorem 1
\[\begin{align} &\text{A continuous function on a bounded closed region } D\subset R^2 \text{ must be bounded, and achieves both its maximum and minimum values.}\\ \end{align}\]
Definition 4: (Partial Derivatives)
\[\begin{align} &\frac{\partial{z}}{\partial x}\bigg|_{(x_0,y_0)}=f_x'(x_0,y_0)=\lim_{\Delta x\rightarrow 0}\frac{\Delta z_x}{\Delta x}=\lim_{\Delta x\rightarrow0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}\\ &f_y'(x_0,y_0)=\lim_{\Delta y\rightarrow0}\frac{\Delta z_y}{\Delta y}=\lim_{y\rightarrow y_0}\frac{f(x_0,y)-f(x_0,y_0)}{y-y_0} \end{align}\]
Examples:
\(\begin{align}
&(1) \text{ Let } f(x,y)=\begin{cases}\frac{xy}{x^2+y^2},&(x,y)\neq(0,0)\\0,&(x,y)=(0,0)\end{cases}. \text{ Find } f_x'(0,0) \text{ and } f'_y(0,0), \text{ though } \lim_{x\rightarrow 0,y\rightarrow 0}f(x,y) \text{ does not exist.}\\
&(2) \text{ Find the partial derivatives of } f(x,y)=\sqrt{x^2+y^2} \text{ at } (0,0), \text{ and analyze its continuity at this point.}\\
\end{align}\)
\(\begin{align}
&\text{(1) } f_x'(0,0)=\lim_{x\rightarrow0}\frac{f(x,0)-f(0,0)}{x-0}=\lim_{x\rightarrow 0}{\frac{0-0}{x}}=0\\
&\text{Similarly, } f_y'(0,0)=0\\
\end{align}\)
Definition 5: (Total Differential)
\[\begin{align} &\text{If } z=f(x,y), \, \Delta z=A\Delta x+B\Delta y+o(\rho) \text{ as } (\rho\rightarrow 0), \text{ where } \rho=\sqrt{\Delta x^2+\Delta y^2}, \text{ then } z=f(x,y) \text{ is said to be}\\ &\text{differentiable at point } P_0, \text{ and its total differential is denoted as:}\\ &dz\bigg|_{(x_0,y_0)}=df\bigg|_{(x_0,y_0)}=A\Delta x+B\Delta y \quad (\text{where } dx = \Delta x, dy = \Delta y) \end{align}\]Notes: \(\begin{align} &(1) \text{ If } \exists A, B \text{ such that } \lim_{\Delta x\rightarrow0,\Delta y\rightarrow 0}\frac{\Delta f-A\Delta x-B\Delta y}{\sqrt{\Delta x^2+\Delta y^2}}=0, \text{ then } f \text{ is differentiable at } (x_0,y_0).\\ &(2) \text{ If } f \text{ is differentiable at } (x_0,y_0), \text{ then } \lim\frac{\Delta f-df}{\sqrt{\Delta x^2+\Delta y^2}}=0\\ \end{align}\)
Theorem 2
\[\begin{align} &\text{If } z=f(x,y) \text{ is differentiable at point } P_0(x_0,y_0), \text{ then its partial derivatives must exist, and:}\\ &dz\bigg|_{(x_0,y_0)}=f'_x(x_0,y_0)dx+f_y'(x_0,y_0)dy\\ \end{align}\]Theorem 3: Relationship Between Statements
\[\begin{align} &\text{Differentiable} \Rightarrow \begin{cases}\text{Continuous} \Rightarrow \text{Limit Exists}\\\text{Partial Derivatives Exist}\\\text{Directional Derivatives Exist}\end{cases}\\ \end{align}\]Examples:
1.
Note: \(\begin{align} &\text{Sufficient Condition for Differentiability: Continuous partial derivative functions at } (x_0,y_0) \Rightarrow \text{Differentiable.}\\ \end{align}\)
2.
Section 2 Methods of Differentiation for Functions of Several Variables
Differentiation of Elementary Functions
Notes: \(\begin{align} &(1) \text{ When differentiating with respect to } x\text{, treat } y \text{ as a constant; when differentiating with respect to } y\text{, treat } x \text{ as a constant.}\\ &(2) u \rightarrow \begin{cases}u_x' \rightarrow \begin{cases}u''_{xx}\\u''_{xy}\end{cases}\\u_y' \rightarrow \begin{cases}u''_{yx}\\u''_{yy}\end{cases}\end{cases}\\ &\text{When the partial derivative functions are continuous, } u''_{xy}=u''_{yx}. \end{align}\)
Examples:
\(\begin{align}
&\text{1. Let } z=\arcsin\frac{x}{\sqrt{x^2+y^2}}, \text{ find } \frac{\partial^2z}{\partial x^2} \text{ and } \frac{\partial^2z}{\partial x\partial y}.\\
&\text{Solution:}\\
&z'_x=\frac{|y|}{x^2+y^2}\\
&z''_{xx}=\frac{\partial}{\partial x}\left(\frac{|y|}{x^2+y^2}\right)=\begin{cases}\frac{-2xy}{(x^2+y^2)^2},&y > 0\\0,&x\neq0,y=0\\\frac{2xy}{(x^2+y^2)^2},&y < 0\end{cases}\\
&z''_{xy}=\frac{\partial}{\partial y}\left(\frac{|y|}{x^2+y^2}\right)=\begin{cases}\frac{x^2-y^2}{(x^2+y^2)^2},&y > 0\\\text{Does not exist},&x\neq 0,y=0\\\frac{y^2-x^2}{(x^2+y^2)^2},&y < 0\end{cases}\\
\end{align}\)
Differentiation of Composite Functions (Chain Rule)
2.
\(\begin{align}
&\frac{\partial z}{\partial x}=-\frac{1}{x^2}f(xy)+\frac{1}{x}f'(xy)y+y\phi'(x+y)\\
&\frac{\partial^2 z}{\partial x\partial y}=-\frac{1}{x^2}f'(xy)x+\frac{1}{x}[f''(xy)xy+f'(xy)]+\phi'(x+y)+y\phi''(x+y)\\
\end{align}\)
3.
\(\begin{align}
&\text{Method 1: Direct Differentiation}\\
&\frac{\partial f}{\partial x}=e^{-(xy)^2}y-e^{-(x+y)^2}\\
&\frac{\partial^2 f}{\partial x\partial y}=\dots\\
&\text{Method 2: Variable Substitution}\\
&\text{Let } u=x+y, v=xy. \text{ Then } f(x,y) \text{ is the composite of } \int_v^ue^{-t^2}dt \text{ with } \begin{cases}u=x+y\\v=xy\end{cases}\\
&\frac{\partial f}{\partial x}=e^{-u^2}\frac{\partial u}{\partial x}-e^{-v^2}\frac{\partial v}{\partial x} = e^{-(x+y)^2}\cdot 1 - e^{-(xy)^2}\cdot y\\
&\frac{\partial^2 f}{\partial x\partial y}=\dots\\
\end{align}\)
Differentiation of Implicit Functions of Several Variables
Example:
\(\begin{align}
&\text{Method 1: Formula Method}\\
&F(x,y,u)=u+e^u-xy, \quad u'_x=\frac{\partial u}{\partial x}=-\frac{F'_x}{F'_u}=-\frac{-y}{1+e^u} = \frac{y}{1+e^u}\\
&\text{Method 2: Implicit Differentiation}\\
&u'_x+e^u u'_x=y \Rightarrow u_x'=\frac{y}{1+e^u}, \quad u'_y={\frac{x}{1+e^u}}\\
&\frac{\partial ^2u}{\partial x\partial y}=\frac{\partial}{\partial y}\left(\frac{y}{1+e^u}\right) = {\frac{1\cdot[1+e^u]-y\cdot e^u\cdot u_y'}{[1+e^u]^2}}\\
\end{align}\)
Methods for Finding Extrema and Absolute Maximum/Minimum Values
Unconstrained Extrema (Two Variables)
\[\begin{align} &(1) \text{ Definition}\\ &(2) \text{ Necessary Conditions: } \begin{cases}z_x'(x_0,y_0)=0\\z_y'(x_0,y_0)=0\end{cases}\\ &(3) \text{ Sufficient Conditions: Let } \Delta=AC-B^2, \text{ where:}\\ &A=f_{xx}''(x_0,y_0), \quad B=f_{xy}''(x_0,y_0), \quad C=f''_{yy}(x_0,y_0)\\ &\begin{cases}\Delta > 0, \text{ Extrema exists } (A<0 \text{ for maximum, } A>0 \text{ for minimum})\\\Delta < 0, \text{ No extrema exists (Saddle point)}\\\Delta = 0, \text{ Inconclusive}\end{cases}\\ \end{align}\]Bounded Closed Regions
\[\begin{align} &\text{Absolute Extrema of a Continuous Function } f(x,y) \text{ on a Bounded Closed Region } D:\\ &\text{If a function } f(x,y) \text{ is continuous on a bounded closed region } D\subset R^2, \text{ then } f(x,y) \text{ must attain both its}\\ &\text{absolute maximum and absolute minimum values within } D.\\ \end{align}\]
Problem Solving Steps
Examples:
