W6-W8. Multivariable Differentiation

Author

Mohammad Alkousa

Published

March 18, 2026

1. Theory

1.1 Functions of Several Variables
1.1.1 Domains, Ranges, and Graphs

Most functions you encounter in calculus depend on a single variable. But in science and engineering, quantities almost always depend on several things at once. The temperature inside a room depends on both the coordinates . The profit of a company depends on dozens of variables. We need a theory for such functions.

A function of several variables is a rule that assigns a unique real number to each point in some subset of . Formally, if is a set of -tuples of real numbers, then a function on assigns to each element of exactly one real number:

The variables are called independent variables, the value is the dependent variable, is the domain, and the set of all output values is the range.

By convention, if no domain is specified, the natural domain is the largest subset of for which the formula gives real numbers.

Finding domains: To find the domain, identify all restrictions:

  • Square roots require the argument to be
  • Logarithms require a strictly positive argument
  • Denominators must be nonzero

Example: For , we need and , so .

The graph of is the set of all points in with and . This is a surface in 3D space.

1.1.2 Interior Points, Boundary Points, Open and Closed Regions

When working with multivariable functions, we need to be precise about the structure of their domains.

Let be a point in a region .

  • is an interior point of if there exists a ball of positive radius centered at that lies entirely within . The collection of all interior points is the interior of . A region is open if it consists entirely of interior points.
  • is a boundary point of if every ball centered at contains both points inside and points outside . A region is closed if it contains all its boundary points.
  • A region is bounded if it fits inside some ball of finite radius; otherwise it is unbounded.
1.1.3 Level Sets

Visualizing a function of two variables as a 3D surface is often difficult. A more practical tool is the level curve.

The set of all points where a function takes a fixed value is called a level set:

  • For : a level set is called a level curve (or contour line).
  • For : it is called a level surface.
  • For : it is called a level hypersurface.

Think of contour lines on a topographic map: each line connects points of equal altitude. Similarly, level curves of connect all points where the function equals .

Example: For , the level curves are circles of radius centered at the origin.

1.2 Limits and Continuity for Functions of Several Variables
1.2.1 The Limit Definition

The concept of limit extends naturally to multiple variables, but with an important twist: in , a point can be approached from infinitely many directions (not just from left or right). This makes limits harder to evaluate — and harder to disprove.

Definition. Let be defined on a domain , and let be a point such that is defined near (but not necessarily at ). Then: means: for every , there exists such that if and , then .

Here, is the Euclidean norm. The key requirement is that the limit value must be the same regardless of the direction along which approaches .

1.2.2 Properties of Limits

These properties mirror those from single-variable calculus. If and , then:

  3. for any
  5. (provided )
  6. for
  7. (for when is even)
  8. Composition: If is continuous at , then

Example: , using the quotient rule and continuity of cosine.

1.2.3 Two-Path Test for Non-existence of a Limit

The most powerful tool for showing a limit does not exist is the Two-Path Test:

If has different limits along two different paths approaching , then does not exist.

The trick is to substitute specific paths (straight lines , parabolas , etc.) and check whether the resulting limits depend on . If they do, the limit does not exist.

Example: Show does not exist.

Along the parabola : This depends on , so the limit does not exist.

1.2.4 Continuity

A function is continuous at the point if:

  1. is defined,
  2. exists,
  3. .

A function is continuous if it is continuous at every point of its domain.

Continuity of compositions: If is continuous at and is continuous at , then the composition is continuous at . This means functions like , , and are continuous on their natural domains.

1.3 Partial Derivatives
1.3.1 Definition

When a function depends on several variables, we can ask: how does it change if we vary only one variable while holding the others fixed? This is the idea of a partial derivative.

Definition. The partial derivative of with respect to at is:

Geometrically, is the slope of the tangent line to the curve obtained by slicing the surface with the vertical plane .

Similarly, the partial derivative with respect to is:

How to compute: To find , treat all variables except as constants and differentiate with respect to using standard rules.

1.3.2 Second and Higher Order Partial Derivatives

Partial derivatives are themselves functions and can be differentiated again. For :

  1. — differentiate with respect to twice
  2. — differentiate with respect to twice
  3. — first , then
  4. — first , then

Clairaut’s (Schwarz’s) Theorem: If , , , , are all continuous at , then . In practice, for smooth functions, the order of differentiation doesn’t matter.

1.3.3 Harmonic Functions and Laplace’s Equation

A function is called harmonic if its second-order partial derivatives exist and satisfy the Laplace equation:

Harmonic functions appear in electrostatics, heat conduction, fluid flow, and many other physical contexts.

Example: is harmonic: , , so .

1.3.4 Differentiability and Linearization

A function is differentiable at if the change can be written as: where , , and as .

Sufficient condition: If and exist and are continuous near , then is differentiable at .

Important implications:

  • Differentiability Continuity (but not vice versa)
  • Existence of partial derivatives alone does NOT guarantee differentiability

The linearization of at is the tangent plane approximation: This is the best linear approximation to near .

1.3.5 Chain Rule

The chain rule extends to functions of several variables.

Case 1: If is differentiable and , are differentiable, then:

Case 2: If and , are all differentiable, then:

General Version: If and each , then:

1.3.6 Implicit Differentiation

If a relation implicitly defines as a differentiable function of , then by the chain rule:

More generally, if defines implicitly, then:

1.4 Directional Derivatives and Gradients
1.4.1 The Directional Derivative

Partial derivatives measure the rate of change along coordinate axes. But what if we want to know how fast changes in an arbitrary direction?

Definition. The directional derivative of at in the direction of a unit vector (where ) is:

This is the rate of change of per unit distance in the direction . If , we get ; if , we get .

Important: The direction vector must be a unit vector. If you’re given a direction vector , first normalize it: .

1.4.2 The Gradient Vector

Definition. The gradient of at a point is the vector of all partial derivatives:

For : .

Key Theorem: If is differentiable at , then:

The directional derivative equals the dot product of the gradient and the direction vector. This elegant formula replaces the limit definition for computation.

1.4.3 Geometric Interpretation of the Gradient

Since (where is the angle between and ):

  • increases most rapidly in the direction of (when ), with rate .
  • decreases most rapidly in the direction of (when ), with rate .
  • has zero rate of change perpendicular to (when ).

Moreover, the gradient is always normal (perpendicular) to level curves (or level surfaces): at any point , is orthogonal to the level curve of through .

1.5 Extreme Values for Functions of Several Variables
1.5.1 Local and Global Extrema

Definition. Let be defined on a region containing . Then:

  • is a local maximum if for all in some open ball around .
  • is a global (absolute) maximum if for all .
  • Similarly for local minimum and global minimum.
1.5.2 Critical Points and the First Derivative Test

Definition. An interior point of the domain of is a critical point if either (all partial derivatives are zero) or at least one partial derivative does not exist at .

First Derivative Test: If has a local extremum at an interior point where partial derivatives exist, then .

Warning: A critical point need not be a local extremum — it could be a saddle point.

Definition. A critical point is a saddle point if every open ball around contains points where and also points where — the function goes up in some directions and down in others, like a mountain pass.

1.5.3 The Second Derivative Test (Hessian)

For , once we find critical points (where ), the Second Derivative Test classifies them.

Define the discriminant (Hessian):

Suppose and all second-order partial derivatives are continuous near . Then:

  1. Local maximum at : if and .
  2. Local minimum at : if and .
  3. Saddle point at : if .
  4. Test inconclusive: if .
1.5.4 Finding a Function from Its Partial Derivatives

Sometimes we know and and need to reconstruct . The method:

  1. Integrate with respect to : , where is an unknown function of .
  2. Differentiate the result with respect to : compare with the given to find .
  3. Integrate to find , then write the final answer.

Alternatively, start by integrating with respect to .

1.6 Gradient Descent Method

One of the most important applications of gradients is the gradient descent method — the backbone of modern machine learning.

1.6.1 The Optimization Problem

We want to solve: where is differentiable.

In practice, this problem may be impossible to solve analytically (the function may be high-dimensional, non-convex, or have many local minima). Numerical algorithms are needed.

1.6.2 The Algorithm

The gradient descent (GD) method generates a sequence of points: starting from an initial point .

The idea is simple: the gradient points in the direction of steepest increase, so moving opposite to it (subtracting ) moves us toward lower values.

Key parameters:

  • is the learning rate (or step size): too large → divergence; too small → very slow convergence.
  • Stopping criteria: (gradient is small), (steps are small), or maximum iterations reached.

Caution: In non-convex problems, gradient descent may converge to a saddle point rather than a local minimum, depending on the starting point .

Example (2D quadratic): Let . Then . Starting from with : The sequence converges to the minimum .

Example (4D quadratic): Let where . Then . The update converges when . With and , each coordinate decays geometrically to .

Convergence in non-convex cases: For non-convex , the algorithm may get trapped at a local minimum or saddle point. Different starting points can lead to different solutions. There is no guarantee of finding the global minimum.

1.7 Constrained Optimization and Lagrange Multipliers
1.7.1 The Problem of Constrained Extrema

Often we want to optimize a function not freely, but subject to constraints. For example: maximize area given a fixed perimeter, or find the point on a surface closest to the origin.

Method of substitution: Sometimes we can express one variable in terms of others using the constraint and substitute, reducing the problem to an unconstrained one. This works well for simple constraints.

1.7.2 Lagrange Multipliers (One Constraint)

When substitution is inconvenient, the method of Lagrange multipliers provides a systematic approach.

Theorem: Suppose and are differentiable and . If is a constrained extremum of subject to , then there exists a scalar (the Lagrange multiplier) such that:

Geometric intuition: At a constrained extremum, the level curves of and the constraint curve are tangent to each other. Being tangent means their normal vectors (the gradients) are parallel, i.e., .

The Lagrangian function:

Setting yields the system:

Practical procedure for subject to : write 4 equations (, , , ) in 4 unknowns () and solve.

1.7.3 Multiple Equality Constraints

For constraints with , use Lagrange multipliers :

The system to solve becomes:

1.7.4 Linear Programming Problem

A linear programming (LP) problem is a special case of constrained optimization where both the objective function and the constraints are linear:

Here the constraints are inequalities rather than equalities. The feasible set (the region satisfying all constraints) is a convex polytope — a polygon in 2D, a polyhedron in 3D, etc.

Key facts:

  • The optimal value of a linear objective over a convex polytope is always attained at a vertex (corner point) of the feasible region.
  • The simplex method systematically moves from vertex to vertex along edges, improving the objective at each step.
  • Lagrange multipliers generalize to KKT (Karush–Kuhn–Tucker) conditions for inequality-constrained problems: at an optimal point , there exist multipliers (one per inequality constraint) such that
  • For LP, the KKT conditions are both necessary and sufficient for global optimality.

Example: Maximize subject to , , , . The feasible region is a polygon; evaluating at all vertices gives the maximum.

1.8 Taylor’s Formula for Functions of Several Variables
1.8.1 Motivation and Derivation

Just as single-variable functions can be approximated by polynomials (Taylor polynomials), multivariable functions can also be approximated by multivariate polynomials. This is useful for error estimation, optimization, and analysis near critical points.

For , parametrize the segment from to as , (where , , ). Define . By the chain rule:

In general, the -th derivative is:

Applying the single-variable Taylor formula to and setting gives:

1.8.2 Taylor’s Formula

Taylor’s Formula for near : where , , and all partial derivatives are evaluated at .

The remainder satisfies:

Special case — quadratic approximation at the origin (when all first derivatives vanish at or as the leading terms):

2. Definitions

  • Function of Several Variables: A rule assigning a unique real number to each point in a domain .
  • Domain: The largest set of inputs for which the function produces real values.
  • Interior Point: A point such that some open ball centered at lies entirely within .
  • Boundary Point: A point such that every ball around it contains both points in and points outside .
  • Open Region: A region consisting entirely of interior points.
  • Closed Region: A region containing all its boundary points.
  • Level Set: The set for a constant .
  • Level Curve: A level set for a function of two variables ().
  • Level Surface: A level set for a function of three variables ().
  • Limit: if approaches from every direction as .
  • Two-Path Test: If has different limits along two paths to the same point, the limit does not exist.
  • Continuity: is continuous at if is defined and .
  • Partial Derivative: ; rate of change of with , holding all other variables fixed.
  • Clairaut’s Theorem: For smooth functions, mixed partial derivatives are equal: .
  • Harmonic Function: A function satisfying the Laplace equation .
  • Differentiable Function: is differentiable at if the change can be well-approximated by its partial derivatives (satisfying the error condition with ).
  • Linearization: The best linear approximation .
  • Gradient: The vector of all partial derivatives.
  • Directional Derivative: ; rate of change in direction (unit vector).
  • Critical Point: An interior point where or some partial derivative does not exist.
  • Saddle Point: A critical point that is neither a local max nor a local min.
  • Hessian / Discriminant: ; used in the Second Derivative Test.
  • Lagrange Multiplier: A scalar such that at a constrained extremum.
  • Lagrangian Function: .
  • Gradient Descent: An iterative algorithm for minimizing .
  • Learning Rate: The step size in gradient descent.
  • Taylor’s Formula (Multivariable): A polynomial approximation of near a point using partial derivatives up to order , plus a remainder term .

3. Formulas

  • Euclidean Norm:
  • Partial Derivative (definition):
  • Chain Rule (1 parameter):
  • Chain Rule (2 parameters): ,
  • Implicit Differentiation ():
  • Implicit Differentiation (): ,
  • Linearization:
  • Gradient:
  • Directional Derivative: (where )
  • Max rate of increase: in direction
  • Hessian / Discriminant:
  • Second Derivative Test: local max if ; local min if ; saddle if
  • Lagrange Condition (1 constraint): and
  • Lagrangian (1 constraint):
  • Lagrangian ( constraints):
  • Gradient Descent:
  • Taylor’s Quadratic (at ): where ,

4. Practice

4.1. Find and Sketch the Domain (Lab 6, Task 1)

Find and sketch the domain for each function:

(a)

(b)

(c)

Click to see the solution

(a) Need : domain is the closed half-plane (on and above the line ).

(b) Need : domain is the open region outside the circle (radius 2).

(c) Need , , : domain is the box .

Answer: (a) ; (b) ; (c)

4.2. Domain Homework (Lab 6, Homework Problem 1)

Find and sketch the domain for each function:

(a)

(b)

Click to see the solution

(a) Need : domain is excluding the circle of radius .

(b) Need : interior of ellipsoid .

Answer: (a) ; (b)

4.3. Sketch Level Curves (Lab 6, Task 2)

Sketch the level curves for:

(a) ,

(b) ,

(c) ,

Click to see the solution

(a) : parallel lines with slope , -intercepts .

(b) : lines through the origin with slopes (undefined at ).

(c) : hyperbolas; for the pair of lines .

Answer: (a) Parallel lines; (b) Lines through origin; (c) Hyperbolas

4.4. Level Curves Homework (Lab 6, Homework Problem 2)

Sketch level curves :

(a) ,

(b) ,

Click to see the solution

(a) : concentric circles of radius centered at origin (a point for ).

(b) : concentric ellipses with semi-axes and (a point for ).

Answer: (a) Concentric circles; (b) Concentric ellipses

4.5. Find Limits (Lab 6, Task 3)

(a)

(b)

(c)

Click to see the solution

(a) As with : (denominator dominates). Answer:

(b) Factor: numerator . Cancel : Answer:

(c) Let : (since ). Answer:

4.6. More Limit Problems — Homework (Lab 6, Homework Problem 3)

(a)

(b)

(c)

(d)

Click to see the solution

(a) Factor denominator: . Cancel : at . Answer:

(b) Let . Expression . Limit does not exist.

(c) Let : . Answer:

(d) Along : . Along : . No limit. Answer: Does not exist

4.7. Show No Limit Exists (Lab 6, Task 4)

Show that the following have no limit as :

(a)

(b)

Click to see the solution

(a) Along -axis (): , giving from right and from left. No limit.

(b) Along -axis (): . Along : . No limit.

Answer: Different paths give different limits in both cases.

4.8. Show No Limit Exists — Homework (Lab 6, Homework Problem 4)

Show that no limit exists as :

(a)

(b) (for )

Click to see the solution

(a) Along : . Along , : , . No limit.

(b) Along : . Along : . Along : . No limit.

Answer: Both functions have different limits along different paths.

4.9. Define for Continuity (Lab 6, Task 5)

Define so that is continuous at the origin.

Click to see the solution

In polar coordinates: as .

Answer:

4.10. Test Continuity — Homework (Lab 6, Homework Problem 5)

Test for continuity at :

(a)

(b)

(c)

Click to see the solution

(a) Polar: . Continuous.

(b) Polar: . Continuous.

(c) Along : . Not continuous.

4.11. Evaluate Limits (Lab 7, Task 1.1)

Evaluate .

Click to see the solution

In polar coordinates: . By L’Hôpital’s rule: .

Answer:

4.12. Evaluate a Limit via Squeeze (Lab 7, Task 1.2)

Evaluate .

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By AM-GM: . So .

Answer:

4.13. Existence of Limit — Three Variables (Lab 7, Task 1.3)

Study the existence of the limit at of .

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The plane passes through the origin, so is undefined on a surface that cuts through every neighborhood of . The limit does not exist in the standard sense because the function is not defined on an entire punctured ball around the origin.

Answer: The limit does not exist (domain excludes the plane through the origin).

4.14. Limit of Three-Variable Function at a Point (Lab 7, Task 1.4)

Study the existence of the limit at of .

Click to see the solution

Along : .

Along (approaching with ): .

Different limits limit does not exist.

Answer: The limit does not exist.

4.15. Compute Partial Derivatives (Lab 7, Task 2.1)

Compute , , , for .

Click to see the solution

Let .

Answer: ; ; ; , where .

4.16. Partial Derivatives with Composite Terms (Lab 7, Task 2.2)

For (), compute and .

Click to see the solution

Answer: ;

4.17. Partial Derivatives of a Power Function (Lab 7, Task 2.3)

For (, ), compute and .

Click to see the solution

Let , so .

Answer: ;

4.18. Check Harmonic Functions (Lab 7, Task 2.4)

Determine whether each function is harmonic:

(a) (b) (c)

Click to see the solution

(a) . Not harmonic.

(b) ; . Sum . Harmonic.

(c) ; . Sum . Harmonic (for ).

4.19. Chain Rule: (Lab 7, Task 2.5)

If , , , compute .

Click to see the solution

Answer:

4.20. Chain Rule: Three-Variable Composition (Lab 7, Task 2.6)

If and , , , compute .

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Substituting , , :

Answer:

4.21. Implicit Differentiation: (Lab 7, Task 2.7)

From , compute .

Click to see the solution

; .

Answer:

4.22. Implicit Differentiation: , (Lab 7, Task 2.8)

If defines , compute and .

Click to see the solution

, , .

Answer: ;

4.23. Directional Derivative (Slope Toward a Point) (Lab 7, Task 2.9)

For , find the slope at in the direction toward .

Click to see the solution

Direction: , .

, .

.

Answer:

4.24. Directional Derivatives — Lab 8 Exercises (Lab 8, Task 1)

Find the slope at point in the direction of on .

Click to see the solution

. .

.

Answer:

4.25. Directional Derivative — Lab 8 Homework (Lab 8, Homework Problem 1)

Determine for in the direction .

Click to see the solution

.

At : . , , .

.

Answer:

4.26. Directional Derivative of (Lab 8, Task 2)

For , find in the direction of .

Click to see the solution

.

; .

.

Answer:

4.27. Classify Critical Points — Lab 8 Homework (Lab 8, Homework Problem 2)

Find and classify the critical points of:

(a)

(b)

Click to see the solution

(a) and .

Substitute: , giving and .

, , , .

  • : → saddle.
  • : , → local min.

(b) ; .

Substitute: . Critical: , , .

.

  • : → saddle.
  • : , → local minima.

Answer:

  1. Saddle at ; local min at .

  2. Saddle at ; local minima at and .

4.28. Directional Derivative at an Angle (Lab 8, Task 3)

Find the directional derivative of in the direction , evaluated at .

Click to see the solution

.

; . At : , .

.

Answer:

4.29. Find Functions from Partial Derivatives (Lab 8, Task 4)

Find for:

(a) ,

(b) ,

(c) ,

Click to see the solution

(a) Integrate w.r.t. : . Diff w.r.t. : . .

(b) Integrate w.r.t. : . Diff w.r.t. : . .

(c) Integrate w.r.t. : . Diff w.r.t. : . .

Answer: (a) ; (b) ; (c)

4.30. Find and Classify Critical Points (Lab 8, Task 5)

Find and classify the critical points of:

(a)

(b)

(c)

Click to see the solution

(a) ; . .

  • : , → local min ().
  • : , → local max ().
  • and : → saddle points.

(b) ; . .

  • : , → local max.
  • : , → local min.
  • and : → saddle points.

(c) and . Critical: , .

  • : saddle (function changes sign nearby).
  • and : local max ().
  • and : local min ().
4.31. Find and Classify Critical Points (Lab 9, Task 1)

Find and classify the critical points of .

Click to see the solution

Key Concept: Use the second derivative test: compute , , set both to zero, then evaluate and at each critical point.

  1. Compute :

  2. Compute :

  3. Set and : Since always, we need:

    • : or .
    • : or .

    Solving simultaneously gives five critical points:

    • , , , .

  4. Compute second partial derivatives:

  5. Classify each critical point:

    (a) At : , , . So . Saddle point at .

    (b) At : Let . . By symmetry, . . , and . Local maximum at . The value is .

    (c) At : By symmetry (), also a local maximum with value .

    (d) At : . , . , . Local minimum at . The value is .

    (e) At : By symmetry, also a local minimum with value .

Answer: Five critical points:

  • : saddle point.
  • (same sign): local maxima with value .
  • (opposite sign): local minima with value .
4.32. Maximum Volume of a Lidless Box (Lab 9, Task 2)

A rectangular box without a lid is to be made from of cardboard. Find the maximum volume.

Click to see the solution

Key Concept: Use Lagrange multipliers. The objective function is the volume ; the constraint is the surface area of a lidless box: the base () plus four sides () equals .

  1. Set up:

    • Objective: .
    • Constraint: .

  2. Lagrange equations (, with ):

  3. Solve by symmetry: Multiply first equation by , second by , third by : So (assuming ).

    From : .

    From and : .

    So .

  4. Apply constraint: , : Thus , .

  5. Maximum volume: .

Answer: The maximum volume is , achieved with base and height .

4.33. Extreme Values on a Circle (Lab 9, Task 3)

Find the extreme values of on the circle .

Click to see the solution

  1. Set up:

    • Objective: .
    • Constraint: .
    • Lagrange function: .

  2. Lagrange equations:

  3. Solve:

    • From : either or .
    • From : either or .

    Case 1: . Then . Constraint: , so . Points: , .

    Case 2: . Then . Constraint: , so . Points: , .

    Case 3: and — impossible on the unit circle.

  4. Classify: Minimum value at ; maximum value at .

Answer: Minimum is (at ); maximum is (at ).

4.34. Points on a Sphere Closest/Farthest from a Point (Lab 9, Task 4)

Find the points on the sphere closest to and farthest from .

Click to see the solution

Key Concept: Minimizing/maximizing the distance is equivalent to minimizing/maximizing the squared distance , which is simpler.

  1. Set up:

    • Objective: .
    • Constraint: .

  2. Lagrange equations: Simplify: , , .

  3. Solve: , , (assuming ). So , , .

    Apply constraint: .

  4. Two solutions:

    • : . Squared distance . Since , this point lies in the same direction as from the origin — closest point.
    • : — farthest point.

    Distance from origin to is . The sphere has radius .

    • Closest distance: .
    • Farthest distance: .

Answer:

  • Closest point: , distance .
  • Farthest point: , distance .
4.35. Maximum and Minimum of on the Unit Circle (Lab 9, Task 5)

Find the maximum and minimum of on the circle .

Click to see the solution
  1. Lagrange equations: , , .
  2. Solve: From the first two equations: and . If : .
    • : . Constraint: . Points: and , .
    • : . Points: and , .

Answer: Maximum is ; minimum is .

4.36. Second Degree Taylor Polynomial at (Lab 9, Task 6)

Find the second degree Taylor polynomial for at .

Click to see the solution

Key Concept: Use the Maclaurin formula , with all derivatives evaluated at .

  1. Compute values at :
    • .
    • .
    • .
    • .
    • .
    • .
  2. Assemble:

Answer: .

4.37. Second Degree Taylor Polynomial at (Lab 9, Task 7)

Find the second degree Taylor polynomial for at .

Click to see the solution
  1. Compute values at :
    • .
    • .
    • .
    • .
    • (since , so ).
    • .
  2. Assemble with , :

Answer: .

4.38. Estimate Using Second Degree Taylor Polynomial (Lab 9, Task 8)

Use a second degree Taylor polynomial to estimate for near .

Click to see the solution

Key Concept: We expand near , then evaluate at , i.e., , .

  1. Compute values at :

    • .
    • .
    • .
    • .
    • .
    • .

  2. Second degree Taylor polynomial: with , :

    • , .
    • , , .
    • .

  3. Estimate:

    (For comparison, the exact value is .)

Answer: .

4.39. Evaluate a Limit by Direct Substitution (Chapter 2, Example 1)

Evaluate .

Click to see the solution

Key Concept: When the denominator is nonzero at the limit point and the outer function is continuous, use the Composition Rule.

  1. Evaluate the inner fraction at :
  2. Apply continuity of cosine: .

Answer:

4.40. Evaluate a Limit Using the Product Rule (Chapter 2, Example 2)

Evaluate .

Click to see the solution
  1. Split using the Product Rule for limits:

Answer:

4.41. Evaluate a Limit by Direct Substitution (3 Variables) (Chapter 2, Example 3)

Evaluate .

Click to see the solution
  1. Substitute directly (the expression is continuous at this point):

Answer:

4.42. Evaluate a Limit Using Factoring (Chapter 2, Example 4)

Evaluate .

Click to see the solution

Key Concept: Factor the numerator to cancel the indeterminate factor .

  1. Factor by grouping:
  2. Cancel and evaluate:

Answer:

4.43. Evaluate a Limit by Rationalizing (Chapter 2, Example 5)

Evaluate .

Click to see the solution

Key Concept: Multiply by the conjugate to eliminate the indeterminate form.

  1. Multiply numerator and denominator by :
  2. Take the limit:

Answer:

4.44. Show a Limit Does Not Exist (Two-Path Test) (Chapter 2, Example 6)

Does the limit exist?

Click to see the solution

Key Concept: If two different paths to give different limits, the limit does not exist.

  1. Along the -axis (, ): .
  2. Along : .

Two different limit values () depending on the path.

Answer: The limit does not exist.

4.45. Show a Limit Does Not Exist (Parabola Paths) (Chapter 2, Example 7)

Show that does not exist.

Click to see the solution
  1. Along (family of parabolas):
  2. This limit depends on : for it equals ; for it equals .

Answer: The limit does not exist.

4.46. Compute Partial Derivatives (Chapter 2, Example 8)

For , find and .

Click to see the solution
  1. (treat as constant):
  2. (treat as constant):

Answer: ;

4.47. Partial Derivatives of a Composite Function (Chapter 2, Example 9)

For , find and .

Click to see the solution
  1. (product rule + chain rule):
  2. (chain rule only, is constant):

Answer: ;

4.48. Implicit Differentiation: One Equation (Chapter 2, Example 10)

Assuming defines implicitly as a function of , find .

Click to see the solution
  1. Define .
  2. Compute and .
  3. Apply: .

Answer:

4.49. Implicit Differentiation: Three Variables (Chapter 2, Example 11)

Assuming defines as a differentiable function of and , find and .

Click to see the solution
  1. , , .
  2. Apply: ; .

Answer: ;

4.50. Chain Rule with One Parameter (Chapter 2, Example 12)

For with and , compute .

Click to see the solution
  1. , ; , .
  2. Chain rule: .
  3. Substitute , and simplify:

Answer:

4.51. Chain Rule with Two Parameters (Chapter 2, Example 13)

For with and , find and .

Click to see the solution
  1. Partial derivatives of :
  2. Partial derivatives of , :
  3. Apply chain rule:

Answer: ;

4.52. Directional Derivative via Gradient (Chapter 2, Example 14)

Find the derivative of at in the direction of .

Click to see the solution
  1. Normalize: .
  2. Gradient at : , .
  3. Directional derivative:

Answer:

4.53. Classify Critical Points (Chapter 2, Example 15)

Find and classify the critical points of .

Click to see the solution
  1. or ; or .
  2. Critical points: , , , .
  3. , :
    • : → saddle point.
    • : , → local maximum.
    • : , → local minimum.
    • : → saddle point.

Answer: Local max at , local min at , saddle points at and .

4.54. Constrained Optimization by Substitution (Chapter 2, Example 16)

Find the point on the plane closest to the origin.

Click to see the solution
  1. Minimize with :
  2. Set : , , so .
  3. Verify minimum: , .

Answer: , distance

4.55. Lagrange Multipliers — One Constraint (Chapter 2, Example 17)

Find the largest and smallest values that takes on the ellipse .

Click to see the solution
  1. Set up with :
  2. Solve: , giving critical points .
  3. Evaluate: (max); (min).

Answer: Maximum ; Minimum .

4.56. Lagrange Multipliers — Two Constraints (Chapter 2, Example 18)

Find the extreme values of subject to and .

Click to see the solution
  1. Set up with multipliers , : , , .
  2. Solve: , giving and .
  3. Evaluate: ; .

Answer: Maximum ; Minimum .

4.57. Taylor Quadratic Approximation (Chapter 2, Example 19)

Find a quadratic approximation to near the origin. How accurate is it for ?

Click to see the solution
  1. Compute derivatives at : ; .
  2. Quadratic Taylor: .
  3. Error bound: Third-order derivatives in absolute value, so

Answer: ; error .

4.58. Lecture Exercises — Chain Rule (Chapter 2, Task 1)

Find :

(1) , ,

(2) , with ,

Click to see the solution

(1)

(2) . Note: is independent of here (the problem only gives and in terms of and ; is treated as a separate variable).

With and :

Answer (1):

4.59. Lecture Exercises — Implicit Differentiation (Chapter 2, Task 2)

Assuming defines as a differentiable function of , find at .

Click to see the solution
  1. .
  2. . At : .
  3. . At : .
  4. .

Answer:

4.60. Lecture Exercises — Gradient (Chapter 2, Task 3)

Find the gradient at the given point:

(1) , at

(2) , at

Click to see the solution

(1) Let .

At : .

At : .

.

(2) At : , so -part: .

Answer (1):

Answer (2):

4.61. Lecture Exercises — Directional Derivative (Chapter 2, Task 4)

Find the derivative at in the direction of :

(1) , ,

(2) , ,

Click to see the solution

(1) Normalize: , .

At : , so .

. At : .

.

(2) Normalize: , .

. At : .

.

Answer (1):

Answer (2):

4.62. Lecture Exercises — Max/Min Directions (Chapter 2, Task 5)

Find the directions of fastest increase and decrease at , and the corresponding rates:

(1) ,

(2) ,

Click to see the solution

(1) At : ; .

, .

Max increase direction: (i.e., ), rate .

Max decrease direction: (i.e., ), rate .

(2) , , . At : , , .

, .

Max increase direction: , rate .

Max decrease direction: , rate .

Answer (1): Max increase: with rate ; max decrease: with rate .

Answer (2): Max increase: with rate ; max decrease: with rate .

4.63. Lecture Exercises — Lagrange Multipliers (Chapter 2, Task 6)

Find the points on the ellipse where has its extreme values.

Click to see the solution
  1. ; Lagrange condition: , .
  2. Solve: , giving , .
  3. Evaluate: .

Answer: Max ; Min .

4.64. Lecture Exercises — Taylor Approximations (Chapter 2, Task 7)

Use Taylor’s formula to find quadratic and cubic approximations of near the origin.

Click to see the solution
  1. Derivatives at :
    • ; ; .
    • ; ; .
    • ; ; ; .
  2. Quadratic approximation ():
  3. Cubic approximation ():

Answer: Quadratic: ; Cubic: