Lecture 40: Differentiation Rules for Multivariable Composite Functions

Chapter Content

• (1) Basic concepts of multivariable functions
• (2) Multivariable Differential Method
• (3) Extrema and extrema of multivariable functions

Section Overview

• (1) Limits of multivariable functions
• (2) Continuity of multivariable functions
• (3) Partial derivative
• (4) Total differential
• (5) Continuous, differentiable, and differentiable relationships

Frequently Tested Question Types in This Section

• Discuss continuity, differentiability, and differentiability

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41.1 Limits of Multivariable Functions

41.1.1 Concept of Multivariable Function Limits

#####Definition: Thelimitofmultivariablefunctions

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

Explanation

• 1. Trends in the limits of multivariable functions:
• In a single variable, there is only one independent variable, so the approach to the target point and the method is relatively simple, only along the X-axis;
• But in binary variables, there are two independent variables, but $(x,y)$ -> $(x_0,y_0)$ is inarbitrary ways
• It must be ensured that in any direction, the function value equals the limit value (similar to the left and right limits in a single-variable limit; only then can this multi-limit be considered to exist;
• 2. Five major characteristics
• (1) Local bounding;
• (2) Number retention;
• (3) Rational operation rules;
• (4) The relationship between limits and infinitesimals;
• (5) Compression;
• Note: L'Hôpital's rule is not available for general multivariable limits.

41.1.2 Basic Methods

Example: Find the value of the limit

• Question: Find $$\lim_{x\to0,\,y\to0}\frac{xy^{2}}{x^{2}+y^{2}}.$$
• Analysis
• This limit has the indeterminate form $0/0$, but L'Hôpital's rule does not apply to a two-variable limit.
• The denominator is quadratic, and the numerator is cubic
• For A/B type:
• If the power of A > the power of B, it is generally 0;
• If the power of A < the power of B, it is generally infinite;
• If the power of A = the power of B, it generally does not exist;
• In this regard, the preliminary judgment is that the limit of the subfunction is 0;
• When in a single element: take the absolute value of x, then expand the -> squeeze theorem;
• Analysis
• Method One:
• $0\leq|\frac{xy^{2}}{x^{2}+y^{2}}|\leq|x|\rightarrow0$
• So $\lim_{x\to0,\,y\to0}\frac{xy^{2}}{x^{2}+y^{2}}=0$
• Method Two:
• You can take the x of the numerator, where x -> 0;
• After the numerator is presented, the remaining parts are not greater than 1 according to their boundedness, so the limit of the entire function is 0;

Example Question: Proving that limits do not exist

• Question: Find $$\lim_{x\to0,\,y\to0}\frac{xy^{}}{x^{2}+y^{2}}$$ Prove that its limit does not exist
• Analysis
• The numerator is 2 times, the denominator is 2 times;
• If there are two different pathpoints and the obtained limit differs, then this limit does not exist;
• Method: Confine y to a single straight line;
• Analysis
• $$\lim_{x\to0,\,y=kx}\frac{xy^{}}{x^{2}+y^{2}} = \lim_{x\to0,\,y\to0}\frac{kx^{2}}{x^{2}+k^{2}x^{2}}=\frac{k}{1+k^2}$$
• Because k is the slope of a straight line, k is not a fixed value. When k takes different slopes, the function values of the limit are different;
• Therefore, this limit does not exist;

41.2 Continuity of Multivariable Functions

Two Concepts

• 1. Concept;
• 2. Discontinuity;
• 3. Nature;

#####Definition: Theconceptofcontinuousmultivariablefunctions

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

#####Theorem: Propertiesofmultivariablecontinuousfunctions

description:

Property 1: The sum, difference, product, and quotient of multivariable continuous functions (with nonzero denominator) are still continuous functions;

Property 2: The composite function of a multivariable continuous function is also a continuous function;

Property 3: Multivariable elementary functions are continuous within theirdefinition region;

Property 4: Maximum Theorem -> A continuous function on a bounded closed region D must find its maximum and minimum values on region D;

Property 5: A continuous function on a bounded closed region D must obtain any value between the maximum and minimum values on region D;

41.3 Partial Derivatives

41.3.1 Basic Concepts of Partial Derivatives

#####Definition: Partialderivative

Description: for $z=f(x,y)$ functions, there are two partial derivatives defined:

1. Partial derivative of X: When partial derivative of X, Y is fixed at$y_0$at (has no effect on x), $x$exist$x.$everywhere$\Delta x$The increment of, at this time there is only one variable x, which is called:

$$\lim_{\Delta x\to0}\frac{f\left(x_{0}+\Delta x,y_{0}\right)-f\left(x_{0},y_{0}\right)}{\Delta x}$$ is the partial derivative with respect to X;

Recorded as:

$\left.\frac{\partial z}{\partial x}\right|_{(x_{0},y_{0})},\left.\frac{\partial f}{\partial x}\right|_{(x_{0},y_{0})},z_{_z}(x_{0},y_{0})\text{ or }$ $f_{x}(x_{0},y_{0})$

2. Partial derivative with respect to Y: The same applies to X,

$$lim_{\Delta y\to0}\frac{f(x_{0},y_{0}+\Delta y)-f(x_{0},y_{0})}{\Delta y}$$

Recorded as:

$\left.\text{}\frac{\partial z}{\partial y}\right|_{(x_{0},y_{0})},\left.\frac{\partial f}{\partial y}\right|_{(x_{0},y_{0})},z_{y}(x_{0},y_{0})\text{ or }f_{y}(x_{0},y_{0}).$

Explanation

• Partial derivative of X:
• Explanation:
• Essentially, it is the derivative of a function of x in one variable, representing the rate of change of the function along the X-axis;
• Other forms: $$\frac{\partial f(x,y)}{\partial x}$$
• Partial derivative of Y:
• Explanation:
• Essentially, it is the derivative of a function of y, representing the rate of change of the function along the Y-axis;
• Other forms:
• $$\frac{\partial f(x,y)}{\partial y}$$
• Summary: Essentially, it is the derivative of a single-variable function;

Method: How to find the partial derivative of x or y

• When $z=f(x,y)$, treat y or x as a constant and take the derivative directly from x or y;
• Example: partial derivative of $z=x^{2}+3xy+y^{2}$ at $(1.2)$;
• Step 1: Calculate the partial derivative function
• Taking partial derivative with respect to x: $\frac{\partial z}{\partial x}=2x+3y$
• Finding partial derivative with respect to y: $$\frac{\partial z}{\partial y}=3x+2y$$
• Step 2: Subtract $(1.2)$ point into the function after finding the partial derivative with respect to x or to y: 2x + 3 y or 3 x + 2 y to get the result

41.3.2 Determining the Existence of a Partial Derivative

Question Type: Determine whether partial derivatives exist

• Method 1: Definition method
• Using the definition of partial derivatives, substitute the function values into the partial derivative of Y and the partial derivative of X, find the limit value of the substituted function, and calculate whether this limit exists;
• Method Two: Essential Method
• Because the partial derivative is essentially the derivative of a function in one variable;
• So you can directly substitute one value of the partial derivative X or Y (the one with the fixed value)
• For example, to find the partial derivative of X, you can directly substitute the value of Y into it to find the limit of the derivative of the function of X;

41.3.3 Geometric Meaning of Partial Derivatives

#####Definition: Thegeometricmeaningofpartialderivativesofabinaryfunction

description: Surface $Z=f(x,y)$

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, $x_0$;

41.3.4 Higher-Order Partial Derivatives

#####Definition: Higherorderpartialderivatives

description:

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

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

Explanation

• Mixed partial derivative:
• $f_{xy}^{\prime\prime}$
• $f_{yx}^{\prime\prime}$

#####Theorem: Higherorderpartialderivativetheorem

description: if the two second-order mixed partial derivatives $\frac{\partial^2z}{\partial x\partial y}$ and $\frac{\partial^2z}{\partial y\partial x}$ of function $z=f(x,y)$ are continuous within region $D$, then within that region:

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

Explanation

• When first x or y is applied first, the limit value is the same;

41.4 Total Differential of Multivariable Functions

#####Definition: Totaldifferentialofmultivariablefunctions

Description: 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 differentiable at point $(x_0,y_0)$;

Multivariate differential: $$dz=A\Delta x+B\Delta y$$

#####Theorem: Differentiablenecessaryconditions

Description: If $z=f(x,y)$ is differentiable at $(x_0,y_0)$, then both partial derivatives exist there and $\mathrm{d}z=\frac{\partial z}{\partial x}\mathrm{d}x+\frac{\partial z}{\partial y}\mathrm{d}y$.

Explanation

• Differentiable -> Differentiable;
• Differentiable -x> Differentiable;
• First-order partial derivative continuous -> differentiable;
• Differentiable -x> First partial derivative continuous;

Method: Use definitions to determine differentiability (sufficient condition)

• (1) Whether $f_{x}(x_{0}, y_{0})$ and $f_{y}(x_{0}, y_{0})$ both exist
• If there is none, then it is certainly non-differentiable;
• If both exist, proceed to (2) for judgment;
• （2）$$\lim_{(\Delta x,\Delta y)\to(0,0)}\frac{\Delta z-[f_{x}(x_{0},y_{0})\Delta x+f_{y}(x_{0,}y_{0})\Delta y]}{\sqrt{\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}}}\text{ Is it zero? }$$
• Explanation:
• According to $\Delta z=f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0)=A\Delta x+B\Delta y+o(\rho)$, place $f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0)$ on the left side;
• See if $\Delta z-(f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0))$ is 0 when divided by $o(\rho)$;
• Conclusion:
• The above limits exist and are 0 -> differentiable;
• If it does not exist or is not 0 -> is not differentiable;
• Meaning:
• The meaning of the above formula -> Because the differential of a point $dz$ equals $A\Delta x+B\Delta y+o(\rho)$, so when the differential of this point exists, whether ${dz-A\Delta x+B\Delta y}\text{ Divide by }{o(\rho)}$ is zero can be seen until its derivative exists;

#####Theorem: Sufficientconditionsfordifferentiability

Description: If the partial derivative $\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$ of $z=f(x,y)$ is continuous at point $(x_0,y_0)$, then function $z=f(x,y)$ is differentiable at point $(x_0,y_0)$;

Explanation

• Concept:
• Two partial derivatives are -> continuously, as a binary function, the partial derivative exists when x and y approach $x_0、y_0$ respectively;
• Conclusion:
• Two partial derivatives -> consecutive differentiable;

41.5 Relationships among Continuous, Partial Derivatives, and Differentiables

Single-variable Function

• Illustration:
• 

Multivariate Functions

• 1. Similarities with the single dollar:
• (1) Continuously -x> differentiable;
• (2) Differentiable -> Deviable;
• (3) Differentiable -> continuous;
• (4) Continuous -x> differentiable;
• 2. Differences from the single yuan:
• Cause:
• The differences are all caused byderivative, so why can't diversity be allowed?
• A partial differentiable is actually the partial derivative of $f_x(x_0,y_0)$ and $f_y(x_0,y_0)$;
• $f_x(x_0,y_0)$ -> $f(x,y_0)$ On the $y=y_0$ line, the function changes only depending on the function value on that line;
• $f_y(x_0,y_0)$ -> $f(x_0,y)$ On the $x=x_0$ line, the function changes only depending on the function value on that line;
• Continuity -> is a heavy limit because it requires that the function is continuous in any direction, while the partial derivative can only determine whether the current derivative exists on a certain line, so it cannot indicate whether it is continuous in any direction;
• Why is differentiable in a single variable: Because in a single variable, the differentiable means that every region is differentiable, so the differentiable -> is differentiable and continuous;
• Concept:
• (1) Deflectable -x> continuity;
• (2) Partially differentiable -x> differentiable;
• 3. New points compared to the One Yuan:
• (1) Partial derivatives are continuous -> differentiable;
• (2) Differentiable -x> Partial Derivatives Continuous;
• Illustration:
• And

41.5 Frequently Tested Question Types

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Question Type: Continuous. Partial Derivatives, Total Derivatives and Their Relationships

PART 1: Problem-solving methods

How to Judge

• 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: $$\lim_{(\Delta x,\Delta y)\to(0,0)}\frac{\Delta z-[f_{x}(x_{0},y_{0})\Delta x+f_{y}(x_{0,}y_{0})\Delta y]}{\sqrt{\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}}}\text{ Is it zero? }$$
• 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): $$\lim_{x\to0,y\to0}f_{x}(x,y)=f_{x}(0,0)$$

PART 2: Typical Example Problems

Example Question: Does the binary function $$f(x,y)=\begin{cases}\frac{xy}{x^{2}+y^{2}},&(x,y)\neq(0,0),\\0,&(x,y)=(0,0)\end{cases}$$ have continuous or partial derivatives at point $(0,0)$?

• Analysis
• Determine whether the function value at this point is continuous -> equal to the function's limit;
• Analysis
• Determining continuity:
• substitute $y=kx$ for $\lim_{x\to0,y\to0}\frac{xy}{x^{2}+y^{2}}$ to prove that its limit does not exist -> discontinuity;
• Determining partial derivatives:
• Find the partial derivative of x: $f_{x}^{\prime}(0,0)=\lim_{x\to0}\frac{f(x,0)-f(0,0)}{x}=\lim_{x\to0}\frac{0-0}{x}$ -> the partial derivative is 0;
• Finding the partial derivative of y: The method is similar to finding x, and the result is also zero partial derivative;
• Therefore, partial derivatives exist;

PART 3: Key Points Review