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Calculus

Comprehensive calculus guide covering limits, derivatives, integrals, applications, and multivariable calculus fundamentals.

calculus derivatives integrals limits optimization differential-equations multivariable

Prerequisites

Required:

  • Pre-Calculus - All topics, especially functions, limits, trigonometry
  • Algebra - Strong foundation in all algebraic manipulation

Helpful:

  • Geometry - For geometric applications and interpretations

Overview

Calculus is the mathematics of change and motion. It provides tools to analyze rates of change (derivatives) and accumulation of quantities (integrals). Calculus is fundamental to physics, engineering, economics, biology, and virtually all quantitative sciences.

This comprehensive guide covers differential calculus (derivatives and their applications), integral calculus (antiderivatives and definite integrals), and introduces multivariable calculus concepts essential for applied mathematics.

Limits

Intuitive Understanding

A limit describes the behavior of a function as its input approaches a particular value.

lim[x→c] f(x) = L means:
"As x gets closer to c, f(x) gets closer to L"

Example: lim[x→2] (x² + 1) = 5
As x approaches 2, x² + 1 approaches 5

Formal Definition (ε-δ Definition)

lim[x→c] f(x) = L if:
For every ε > 0, there exists δ > 0 such that
if 0 < |x - c| < δ, then |f(x) - L| < ε

This makes precise the idea of "getting arbitrarily close"

Types of Limits

One-Sided Limits

Right-hand limit: lim[x→c⁺] f(x)
Left-hand limit: lim[x→c⁻] f(x)

Example: f(x) = |x|/x
lim[x→0⁺] |x|/x = 1 (from right)
lim[x→0⁻] |x|/x = -1 (from left)

Since one-sided limits differ, lim[x→0] |x|/x does not exist

Limits at Infinity

Horizontal asymptotes:
lim[x→∞] f(x) = L means y = L is a horizontal asymptote

Example: lim[x→∞] 1/x = 0
Example: lim[x→∞] (3x² + 2x)/(x² - 1) = 3

Infinite Limits

Vertical asymptotes:
lim[x→c] f(x) = ∞ means f(x) grows without bound as x → c

Example: lim[x→0⁺] 1/x = ∞
Example: lim[x→2] 1/(x-2)² = ∞

Limit Laws

If lim[x→c] f(x) = L and lim[x→c] g(x) = M, then:

1. lim[x→c] [f(x) + g(x)] = L + M
2. lim[x→c] [f(x) - g(x)] = L - M  
3. lim[x→c] [f(x) · g(x)] = L · M
4. lim[x→c] [f(x)/g(x)] = L/M (if M ≠ 0)
5. lim[x→c] [f(x)]ⁿ = Lⁿ
6. lim[x→c] ⁿ√[f(x)] = ⁿ√L (if appropriate)

Evaluating Limits

Direct Substitution

If f is continuous at c, then lim[x→c] f(x) = f(c)

Example: lim[x→3] (x² + 2x - 1) = 9 + 6 - 1 = 14

Factoring

For rational functions with 0/0 form:

Example: lim[x→2] (x² - 4)/(x - 2)
= lim[x→2] [(x-2)(x+2)]/(x-2)
= lim[x→2] (x + 2) = 4

Rationalization

For expressions with radicals:

Example: lim[x→0] (√(x+1) - 1)/x
Multiply by (√(x+1) + 1)/(√(x+1) + 1):
= lim[x→0] [(x+1) - 1]/[x(√(x+1) + 1)]
= lim[x→0] x/[x(√(x+1) + 1)]
= lim[x→0] 1/(√(x+1) + 1) = 1/2

L’Hôpital’s Rule

For indeterminate forms 0/0 or ∞/∞:
lim[x→c] f(x)/g(x) = lim[x→c] f'(x)/g'(x)

Example: lim[x→0] sin(x)/x
= lim[x→0] cos(x)/1 = 1

Special Limits

Important limits to remember:
lim[x→0] sin(x)/x = 1
lim[x→0] (1 - cos(x))/x = 0
lim[x→0] (1 + x)^(1/x) = e
lim[x→∞] (1 + 1/x)^x = e
lim[x→0] (e^x - 1)/x = 1
lim[x→0] ln(1 + x)/x = 1

Derivatives

Definition

The derivative of f at x is the instantaneous rate of change:

f'(x) = lim[h→0] [f(x+h) - f(x)]/h

Geometric interpretation: slope of tangent line
Physical interpretation: instantaneous velocity

Basic Derivative Rules

Power Rule

d/dx [x^n] = nx^(n-1)

Examples:
d/dx [x³] = 3x²
d/dx [x^(-2)] = -2x^(-3)
d/dx [√x] = d/dx [x^(1/2)] = (1/2)x^(-1/2) = 1/(2√x)

Constant and Linear Rules

d/dx [c] = 0 (constant)
d/dx [cx] = c (constant multiple)
d/dx [f(x) + g(x)] = f'(x) + g'(x) (sum)
d/dx [f(x) - g(x)] = f'(x) - g'(x) (difference)

Advanced Derivative Rules

Product Rule

d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

Example: d/dx [x²sin(x)]
= 2x·sin(x) + x²·cos(x)
= 2x sin(x) + x² cos(x)

Quotient Rule

d/dx [f(x)/g(x)] = [f'(x)g(x) - f(x)g'(x)]/[g(x)]²

Example: d/dx [sin(x)/x²]
= [cos(x)·x² - sin(x)·2x]/(x²)²
= [x²cos(x) - 2x sin(x)]/x⁴
= [x cos(x) - 2 sin(x)]/x³

Chain Rule

d/dx [f(g(x))] = f'(g(x))·g'(x)

Example: d/dx [sin(x²)]
= cos(x²)·2x = 2x cos(x²)

Example: d/dx [(3x + 1)⁵]
= 5(3x + 1)⁴·3 = 15(3x + 1)⁴

Derivatives of Special Functions

Trigonometric Functions

d/dx [sin(x)] = cos(x)
d/dx [cos(x)] = -sin(x)
d/dx [tan(x)] = sec²(x)
d/dx [sec(x)] = sec(x)tan(x)
d/dx [csc(x)] = -csc(x)cot(x)
d/dx [cot(x)] = -csc²(x)

Exponential and Logarithmic Functions

d/dx [e^x] = e^x
d/dx [a^x] = a^x ln(a)
d/dx [ln(x)] = 1/x
d/dx [log_a(x)] = 1/(x ln(a))

Inverse Trigonometric Functions

d/dx [arcsin(x)] = 1/√(1-x²)
d/dx [arccos(x)] = -1/√(1-x²)
d/dx [arctan(x)] = 1/(1+x²)

Implicit Differentiation

When y is defined implicitly by an equation F(x,y) = 0:

Example: x² + y² = 25
Differentiate both sides:
2x + 2y(dy/dx) = 0
Solve for dy/dx:
dy/dx = -x/y

Example: xy + sin(y) = 1
d/dx[xy] + d/dx[sin(y)] = d/dx[1]
y + x(dy/dx) + cos(y)(dy/dx) = 0
dy/dx = -y/(x + cos(y))

Higher-Order Derivatives

f'(x) or f^(1)(x) = first derivative
f''(x) or f^(2)(x) = second derivative
f'''(x) or f^(3)(x) = third derivative
f^(n)(x) = nth derivative

Example: f(x) = x⁴
f'(x) = 4x³
f''(x) = 12x²
f'''(x) = 24x
f^(4)(x) = 24
f^(5)(x) = 0

Applications of Derivatives

When quantities are related by an equation, their rates of change are related by the derivative of that equation.

Example: Balloon Problem
A spherical balloon's radius increases at 2 cm/min. 
How fast is volume increasing when r = 10 cm?

Given: dr/dt = 2 cm/min
Find: dV/dt when r = 10

Volume formula: V = (4/3)πr³
Differentiate: dV/dt = 4πr²(dr/dt)
Substitute: dV/dt = 4π(10)²(2) = 800π cm³/min

Optimization Problems

Find maximum or minimum values using calculus.

Process:

  1. Define variables and constraints
  2. Express the quantity to optimize as a function
  3. Find critical points (where f’(x) = 0)
  4. Use first or second derivative test
  5. Check endpoints if applicable
Example: Fence Problem
Farmer has 1000 feet of fence to enclose rectangular area.
What dimensions maximize the area?

Let x = width, y = length
Constraint: 2x + 2y = 1000, so y = 500 - x
Area: A(x) = xy = x(500 - x) = 500x - x²

Find critical points:
A'(x) = 500 - 2x = 0
x = 250 feet

Therefore: y = 500 - 250 = 250 feet
Maximum area = 250 × 250 = 62,500 square feet

Linear Approximation and Differentials

Linear approximation (tangent line approximation):
f(x) ≈ f(a) + f'(a)(x - a)

Example: Approximate √26
Let f(x) = √x, a = 25
f'(x) = 1/(2√x), so f'(25) = 1/10

√26 ≈ √25 + (1/10)(26 - 25) = 5 + 0.1 = 5.1
(Actual value: √26 ≈ 5.099...)

Differential: dy = f'(x)dx represents change in y

Motion Problems

If position is s(t):
• Velocity: v(t) = s'(t)
• Acceleration: a(t) = v'(t) = s''(t)
• Speed: |v(t)|

Example: s(t) = t³ - 6t² + 9t + 1
v(t) = 3t² - 12t + 9
a(t) = 6t - 12

When is particle at rest? v(t) = 0
3t² - 12t + 9 = 0
3(t² - 4t + 3) = 0
3(t - 1)(t - 3) = 0
At rest when t = 1 and t = 3

Curve Sketching

Use derivatives to analyze function behavior:

First Derivative Test

• f'(x) > 0: function increasing
• f'(x) < 0: function decreasing
• f'(x) = 0: critical points (potential extrema)
• f'(x) changes sign: local maximum or minimum

Second Derivative Test

• f''(x) > 0: concave up (cup shape)
• f''(x) < 0: concave down (cap shape)
• f''(x) = 0: potential inflection point
• f''(c) = 0 and f''(x) changes sign at c: inflection point

For critical point c where f'(c) = 0:
• f''(c) > 0: local minimum
• f''(c) < 0: local maximum
• f''(c) = 0: test is inconclusive

Integrals

Antiderivatives

An antiderivative of f(x) is a function F(x) such that F’(x) = f(x).

∫ f(x) dx = F(x) + C

where C is the constant of integration

Example: ∫ 2x dx = x² + C
Verification: d/dx[x² + C] = 2x ✓

Basic Integration Rules

Power Rule

∫ x^n dx = x^(n+1)/(n+1) + C  (n ≠ -1)

Examples:
∫ x³ dx = x⁴/4 + C
∫ √x dx = ∫ x^(1/2) dx = (2/3)x^(3/2) + C
∫ 1/x² dx = ∫ x^(-2) dx = -x^(-1) + C = -1/x + C

Basic Integrals

∫ k dx = kx + C  (constant)
∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
∫ kf(x) dx = k∫ f(x) dx

∫ e^x dx = e^x + C
∫ 1/x dx = ln|x| + C
∫ sin(x) dx = -cos(x) + C
∫ cos(x) dx = sin(x) + C
∫ sec²(x) dx = tan(x) + C

Integration Techniques

Substitution (u-substitution)

∫ f(g(x))g'(x) dx = ∫ f(u) du  where u = g(x)

Example: ∫ 2x cos(x²) dx
Let u = x², then du = 2x dx
∫ cos(u) du = sin(u) + C = sin(x²) + C

Example: ∫ (3x + 1)⁴ dx
Let u = 3x + 1, then du = 3 dx, so dx = du/3
∫ u⁴ · (du/3) = (1/3) ∫ u⁴ du = (1/3) · u⁵/5 + C = (3x + 1)⁵/15 + C

Integration by Parts

∫ u dv = uv - ∫ v du

Choose u and dv so that ∫ v du is simpler than ∫ u dv

Example: ∫ x e^x dx
Let u = x, dv = e^x dx
Then du = dx, v = e^x
∫ x e^x dx = x·e^x - ∫ e^x dx = xe^x - e^x + C = e^x(x - 1) + C

Definite Integrals

Fundamental Theorem of Calculus

Part 1: If f is continuous on [a,b] and F'(x) = f(x), then:
∫[a to b] f(x) dx = F(b) - F(a)

Part 2: If f is continuous on [a,b], then:
d/dx ∫[a to x] f(t) dt = f(x)

Evaluation

∫[a to b] f(x) dx = [F(x)][a to b] = F(b) - F(a)

Example: ∫[1 to 4] x² dx
= [x³/3][1 to 4] = 64/3 - 1/3 = 63/3 = 21

Properties

∫[a to a] f(x) dx = 0
∫[a to b] f(x) dx = -∫[b to a] f(x) dx
∫[a to b] f(x) dx + ∫[b to c] f(x) dx = ∫[a to c] f(x) dx
∫[a to b] [f(x) + g(x)] dx = ∫[a to b] f(x) dx + ∫[a to b] g(x) dx
∫[a to b] kf(x) dx = k∫[a to b] f(x) dx

Applications of Integration

Area Under Curves

Area between f(x) and x-axis from a to b:
A = ∫[a to b] |f(x)| dx

Area between curves f(x) and g(x):
A = ∫[a to b] |f(x) - g(x)| dx

Example: Area between y = x² and y = 4
Intersection points: x² = 4, so x = ±2
A = ∫[-2 to 2] (4 - x²) dx = [4x - x³/3][-2 to 2] = 32/3

Volume of Revolution

Disk Method:

V = π∫[a to b] [f(x)]² dx

Example: Volume when y = √x is revolved around x-axis from x = 0 to x = 4
V = π∫[0 to 4] (√x)² dx = π∫[0 to 4] x dx = π[x²/2][0 to 4] = 8π

Washer Method:

V = π∫[a to b] ([R(x)]² - [r(x)]²) dx

where R(x) is outer radius, r(x) is inner radius

Average Value

Average value of f on [a,b]:
f_avg = (1/(b-a)) ∫[a to b] f(x) dx

Example: Average value of f(x) = x² on [0,3]
f_avg = (1/3) ∫[0 to 3] x² dx = (1/3) · [x³/3][0 to 3] = (1/3) · 9 = 3

Multivariable Calculus Introduction

Functions of Several Variables

z = f(x,y) represents a surface in 3D space

Examples:
f(x,y) = x² + y² (paraboloid)
f(x,y) = sin(x)cos(y) (wave surface)
f(x,y) = √(9 - x² - y²) (upper hemisphere)

Partial Derivatives

∂f/∂x: derivative with respect to x, treating y as constant
∂f/∂y: derivative with respect to y, treating x as constant

Example: f(x,y) = x²y + 3xy²
∂f/∂x = 2xy + 3y²
∂f/∂y = x² + 6xy

Geometric interpretation:
∂f/∂x gives slope in x-direction
∂f/∂y gives slope in y-direction

Multiple Integrals

Double integral: ∫∫[R] f(x,y) dA

Represents volume under surface z = f(x,y) over region R

Example: ∫[0 to 2] ∫[0 to 1] (x + y) dy dx
= ∫[0 to 2] [xy + y²/2][0 to 1] dx
= ∫[0 to 2] (x + 1/2) dx
= [x²/2 + x/2][0 to 2] = 2 + 1 = 3

Practice Problems

Limits

1. lim[x→3] (x² - 9)/(x - 3)
2. lim[x→0] sin(5x)/x
3. lim[x→∞] (2x² + 3x)/(x² - 1)

Derivatives

1. Find dy/dx: y = x³ sin(x)
2. Find dy/dx using implicit differentiation: x² + xy + y² = 7
3. Find the equation of the tangent line to y = e^x at x = 0

Applications

1. A ladder 25 feet long leans against a wall. The bottom slides away at 3 ft/sec. How fast is the top sliding down when the bottom is 15 feet from the wall?

2. Find the maximum area of a rectangle inscribed in the ellipse x²/9 + y²/4 = 1

3. Find the critical points of f(x) = x³ - 3x² - 9x + 5

Integration

1. ∫ (3x² - 2x + 1) dx
2. ∫ x cos(x²) dx
3. ∫[0 to π/2] sin(x) dx
4. Find the area between y = x² and y = 2x

Answer Key

Limits

1. lim[x→3] (x² - 9)/(x - 3) = lim[x→3] (x + 3) = 6
2. lim[x→0] sin(5x)/x = 5·lim[x→0] sin(5x)/(5x) = 5·1 = 5
3. lim[x→∞] (2x² + 3x)/(x² - 1) = 2

Derivatives

1. dy/dx = 3x² sin(x) + x³ cos(x)
2. 2x + y + x(dy/dx) + 2y(dy/dx) = 0
   dy/dx = -(2x + y)/(x + 2y)
3. y = e^x, y' = e^x, at x = 0: point (0,1), slope = 1
   Tangent line: y = x + 1

Applications

1. Using Pythagorean theorem: x² + y² = 625
   2x(dx/dt) + 2y(dy/dt) = 0
   When x = 15, y = 20, dx/dt = 3
   dy/dt = -xy(dx/dt)/y = -15(3)/20 = -2.25 ft/sec

2. Rectangle vertices at (±x, ±y) where x²/9 + y²/4 = 1
   Area A = 4xy, y = 2√(1 - x²/9)
   A(x) = 8x√(1 - x²/9), A'(x) = 0 gives x = 3/√2
   Maximum area = 12

3. f'(x) = 3x² - 6x - 9 = 3(x² - 2x - 3) = 3(x - 3)(x + 1)
   Critical points: x = -1, x = 3

Integration

1. ∫ (3x² - 2x + 1) dx = x³ - x² + x + C
2. ∫ x cos(x²) dx = (1/2) sin(x²) + C
3. ∫[0 to π/2] sin(x) dx = [-cos(x)][0 to π/2] = 1
4. Intersection: x² = 2x gives x = 0, 2
   Area = ∫[0 to 2] (2x - x²) dx = 4/3

Quick Reference

Derivative Rules

  • Power Rule: d/dx[x^n] = nx^(n-1)
  • Product Rule: d/dx[fg] = f’g + fg'
  • Quotient Rule: d/dx[f/g] = (f’g - fg’)/g²
  • Chain Rule: d/dx[f(g(x))] = f’(g(x))·g’(x)

Integration Rules

  • Power Rule: ∫x^n dx = x^(n+1)/(n+1) + C
  • ∫e^x dx = e^x + C
  • ∫sin(x) dx = -cos(x) + C
  • ∫cos(x) dx = sin(x) + C

Fundamental Theorem

  • ∫[a to b] f(x) dx = F(b) - F(a) where F’(x) = f(x)

See Also

Categories:
math
Last updated: January 1, 2023