Chapter 7: Multivariable Differential Calculus

7.1 Fundamentals of Multivariable Functions

Definition: Limits of multivariable functions

$$\lim_{(x,y)\to(x_0,y_0)}f(x,y)=A$$

Definition: Continuity of multivariable functions

$$\lim_{(x,y)\to(x_0,y_0)}f(x,y)=f(x_0,y_0)$$

Definition: Partial derivative

Against X:
$$\lim_{\Delta x\to0}\frac{f\left(x_{0}+\Delta x,y_{0}\right)-f\left(x_{0},y_{0}\right)}{\Delta x}$$
Other forms: $$\frac{\partial f(x,y)}{\partial x}$$
For Y:
$$lim_{\Delta y\to0}\frac{f(x_{0},y_{0}+\Delta y)-f(x_{0},y_{0})}{\Delta y}$$
Other forms:
$$\frac{\partial f(x,y)}{\partial y}$$

Definition: The geometric meaning of partial derivatives

Curved surfaces $Z=f(x,y)$
1. $f_x(x_0,y_0)$ represents the tangent line to $y=f(x)$ on the $y=f(x)$ line, $x_0$;
2. $f_y(x_0,y_0)$ represents the tangent line to $x=f(y)$ on the $x=f(y)$ line, $x_0$;

Definition: Higher-order partial derivative

$$\frac{\partial}{\partial x}\biggl(\frac{\partial z}{\partial x}\biggr)=\frac{\partial^{2}z}{\partial x^{2}}=f_{xx}^{\prime\prime}\quad\frac{\partial}{\partial y}\biggl(\frac{\partial z}{\partial x}\biggr)=\frac{\partial^{2}z}{\partial x\partial y}=f_{xy}^{\prime\prime}$$
$$\frac{\partial}{\partial x}\Bigg(\frac{\partial z}{\partial y}\Bigg)=\frac{\partial^{2}z}{\partial y\partial x}=f_{yx}^{\prime\prime}\quad\frac{\partial}{\partial y}\Bigg(\frac{\partial z}{\partial y}\Bigg)=\frac{\partial^{2}z}{\partial y^{2}}=f_{yy}^{\prime\prime}$$

And: $$\frac{\partial^2z}{\partial x\partial y}=\frac{\partial^2z}{\partial y\partial x}$$

Definition: The total differential of a multivariable function

If $\Delta z=f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0)=A\Delta x+B\Delta y+o(\rho)$, then the function $z=f(x,y)$ is called differentiable at point $(x_0,y_0)$;
That is: $$dz=A\Delta x+B\Delta y$$

Theorem: A necessary condition for differentiability

If $z=f(x,y)$ is differentiable at $(x_0,y_0)$, then both partial derivatives exist there and $\mathrm{d}z=\frac{\partial z}{\partial x}\mathrm{d}x+\frac{\partial z}{\partial y}\mathrm{d}y$.
Differentiable -> Differentiable;
Differentiable -x> Differentiable;
First-order partial derivative continuous -> differentiable;
Differentiable -x> First partial derivative continuous;

Theorem: A sufficient condition for differentiability

If both first-order partial derivatives of $z=f(x,y)$ are continuous at $(x_0,y_0)$, then $z$ is differentiable at that point.

Problem Solving: Judgments of differentiable, differentiable, continuous, and partial differentiations of continuity

Determining whether differentiable:
1. Definition:
Step 1: Determine whether $f_{x}(x_{0}, y_{0})$ and $f_{y}(x_{0}, y_{0})$ both exist;
Step 2: Use the following definitions to determine whether a double limit exists;
Formula: $$\lim_{(\Delta x,\Delta y)\to(0,0)}\frac{\Delta z-[f_{x}(x_{0},y_{0})\Delta x+f_{y}(x_{0,}y_{0})\Delta y]}{\sqrt{\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}}}$$
Determining whether it is differentiable:
1. Definition method: Find the partial derivative of x and y separately, use partial derivatives to define and analyze whether the univariate function derivative corresponding to the partial derivative exists;
Formula (X partial derivative): $$\lim_{\Delta x\to0}\frac{f\left(x_{0}+\Delta x,y_{0}\right)-f\left(x_{0},y_{0}\right)}{\Delta x}$$
2. Bring first, ask later:
Determining continuity:
1. Definition:
(1) Prove continuity: Whether the function value at a certain point equals the limit value of that function; if equal, then continuous;
(2) Proving discontinuity: same;
2. Special y Value Method:
Substitute a fixed y function into a function, often used to prove discontinuity;
Common Forms:
$y=kx$
$y=0$
Determining whether the partial derivative is continuous:
Note:
Multivariable function continuous ≠ Multivariable function partial derivative continuous;
Whether the partial derivative is continuous refers to whether thepartial derivative of the derivative with respect to X or Yis continuous;
1. Definition method: Use partial derivative functions with respect to X or Y to solve:
If the following formula does not hold, the partial derivative is discontinuous;
Formula (partial derivative function with respect to X): $$\lim_{x\to0,y\to0}f_{x}(x,y)=f_{x}(0,0)$$

7.2 Multivariable Differential Methods

Problems in multivariable function differentiation:

(1) Differential method for composite functions;
(2) Implicit function differentiation;

Theorem: The differentiation rule for multivariable composite functions

Let $u=u(x,y)$ and $v=v(x,y)$ have first-order partial derivatives, and let $z=f(u,v)$ have continuous first-order partial derivatives. Then the composite function $z=f(u(x,y),v(x,y))$ has:
$$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x},\qquad\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$$

Theorem: Invariance of the form of the whole differential

Let functions $z=f(u,v),\quad u=u(x,v)$ and $\nu=\nu(x,y)$ have continuous first-order partial derivatives, then the total differential invariance of the composite function $z=f[u(x,y),v(x,y)]$:
$$\mathrm{d}z=\frac{\partial z}{\partial x}\mathrm{d}x+\frac{\partial z}{\partial y}\mathrm{d}y=\frac{\partial z}{\partial u}\mathrm{d}u+\frac{\partial z}{\partial v}\mathrm{d}v.$$
Multivariable functions also have the invariance of differential forms;
Therefore: $$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x},\qquad\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}.$$

Definition: Multivariable function implicit function

The implicit function $y=(x)$ determined by equation $F(x, y)=0$ gives:
$$y^{\prime}=-\frac{F_{x}^{\prime}}{F_{y}^{\prime}}$$

Theorem: The existence theorem of implicit functions

1. $F(x,y)$ Continuously partial differentiable within the neighborhood of point $M_0(x_0,y_0)$;
2. $F(x_0,y_0)=0$；
3. $F_y\:\prime(x_0,y_0)\neq0$
Then function $F(x,y)$ in the $M_{0}$ neighborhood uniquely determines a continuous differentiable function $y=f(x)$ to $y_0=f(x_0)$
then: $$\frac{dy}{dx}=-\frac{F_{x}'}{F_{y}'}$$

Theorem: The existence of a theorem for multivariable implicit functions

$F(x,y,z)$ is continuously differentiable in the $M_{0}(x_{0},y_{0},z_{0})$ neighborhood, and $F(x_{0},y_{0},z_{0})=0,F_{z}\:{\prime}(x_{0},y_{0},z_{0})\neq0$ is uniquely determined by $F(x,y,z)=0$ in the $M_{0}$ neighborhood of a continuous differentiable function $z=\varphi(x,y)$ and $z_0=\varphi(x_0,y_0)$
Therefore, $$\frac{\partial z}{\partial x}=-\frac{F_{x}^{\prime}}{F_{z}^{\prime}},\qquad\frac{\partial z}{\partial y}=-\frac{F_{y}^{\prime}}{F_{z}^{\prime}}.$$

7.3 Extrema and Extrema of Multivariable Functions

Definition: Extremum of multivariable functions

In a neighborhood, the maximum or minimum value of the multivariable function;
If the inequality always holds in a neighborhood of point $(x_0,y_0)$: $$f(x,y)\leq f(x_0,y_0)\quad(f(x,y)\geq f(x_0,y_0))$$
Then $f$ has a local maximum (minimum) at $(x_0,y_0)$, and $(x_0,y_0)$ is called a local maximum (minimum) point.
The maximum and minimum are collectively called extremums, while the maximum and minimum points are collectively called extremum points;

Definition: The dwell point of a multivariable function

The point where the partial derivative of a multivariable function equals zero is called the stationary point of the multivariable function;

Theorem: A necessary condition for the existence of extrema in multivariable functions

If $z=f(x,y)$ has a partial derivative at point $(x_0,y_0)$ and takes an extremum at $(x_0,y_0)$, then there is: $$f_{x}^{\prime}(x_{0},y_{0})=0,\,f_{y}^{\prime}(x_{0},y_{0})=0$$

Theorem: Sufficient condition for an extremum

Determining the Extreme Value of Static Points in Multivariable Functions

When $z=f(x,y)$ has a second-order continuous partial derivative in a neighborhood of point $P_0(x_0,y_0)$, and $f_x^{\prime}(x_0,y_{0})=f_y^{\prime}(x_0,y_{0})=0$, $A=f_{xx}^{\prime\prime}(x_{0},y_{0})\quad\quad B=f_{xy}^{\prime\prime}(x_{0},y_{0})\quad\quad C=f_{yy}^{\prime\prime}(x_{0},y_{0})$;
1. If it is $AC-B^{2}>0$, then there are extremes
1. A<0 is the maximum;
1. A>0 is the minimum value;
1. If it is $AC-B^{2}<0$, then there is no extreme value
2. If it is $AC-B^{2}=0$, then it can only be determined by definition

Definition: Conditional extremum

Given $z=f(x,y)$, its extremum is under the condition $\varphi(x,y)=0$, find its extremum as the conditional extremum;

Theorem: Lagrange multiplier method

(1) For a single constraint: let $F(x,y,\lambda)=f(x,y)+\lambda\varphi(x,y)$
Then the necessary condition for obtaining the extremum is: $$\begin{cases}F_x=f_x^{\prime}(x,y)+\lambda\varphi_x^{\prime}(x,y)=0,\\F_y=f_y^{\prime}(x,y)+\lambda\varphi_y^{\prime}(x,y)=0,\\F_\lambda=\varphi(x,y)=0,\end{cases}$$
(2) When multiple constraints: the extremum of the condition under $f(x,y,z)\text{ In terms of conditions }\varphi(x,y,z)=0,\psi(x,y,z)=0$
Ling $F(x,y,z,\lambda,\mu)=f(x,y,z)+\lambda\varphi(x,y,z)+\mu\psi(x,y,z)$