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Definition
Formula Procedure Example f'(x) = d/dx (3x^2) f'(x) = 2 * 3 * x^(2-1) f'(x) = 6x So, f'(x) = 6x is the derivative of f(x) = 3x^2. 2 Definition
Formula
Procedure Example Let u(x) = x^2 and v(x) = sin(x). Apply the product rule: [u * v]' = u'v + uv' u' = 2x (derivative of x^2) v' = cos(x) (derivative of sin(x)) [x^2 * sin(x)]' = (2x * sin(x)) + (x^2 * cos(x))
Derivative (Instantaneous Rate of Change) The derivative of a function at a point represents the instantaneous rate of change or slope of the function at that point.
Apply the appropriate rule(s) to find the derivative of a given function. Use the product rule to find the derivative of f(x) = x^2 * sin(x).
The derivative of a function f(x) is denoted as f'(x) or dy/dx. Differentiate the function using differentiation rules (power rule, product rule, chain rule, etc.).Find the derivative of f(x) = 3x^2.
Differentiation Rules Differentiation rules are a set of guidelines and formulas for finding derivatives of various functions. Various rules include the power rule, product rule, quotient rule, chain rule, and more.
Definition • Formula • Procedure • Examples & Solutions
STUDY GUIDE
Mathematics: CALCULUS
DIFFERENTIAL
CALCULUS
Definition
Formula Procedure
Example
Definition
Formula Procedure Example Let u(x) = x^2 and v(x) = e^x. Apply the product rule: [u * v]' = u'v + uv' u' = 2x (derivative of x^2) v' = e^x (derivative of e^x) [x^2 * e^x]' = (2x * e^x) + (x^2 * e^x) 5 Definition
Formula Procedure Example Apply the quotient rule: [u/v]' = (u'v - uv') / (v^2) Let u(x) = x^2 + 1 and v(x) = x. u' = 2x (derivative of x^2 + 1) v' = 1 (derivative of x) [ (x^2 + 1) / x ]' = [(2x * x) - ((x^2 + 1) * 1)] / (x^2) 6 Definition
Formula Procedure
Power Rule The power rule is a differentiation rule used to find the derivative of a function in the form of x^n. If f(x) = x^n, then f'(x) = nx^(n-1). Apply the power rule to functions with a single term raised to a constant exponent.
Find the derivative of f(x) = x^2 * e^x.
Product Rule The product rule is a differentiation rule used to find the derivative of the product of two functions. If f(x) = u(x) * v(x), then f'(x) = u'v + uv'. Apply the product rule to functions that are products of two functions of x.
Find the derivative of f(x) = 5x^3. f'(x) = d/dx (5x^3) f'(x) = 3 * 5 * x^(3-1) f'(x) = 15x^ So, f'(x) = 15x^2 is the derivative of f(x) = 5x^
Chain Rule The chain rule is a differentiation rule used to find the derivative of a composition of functions. If y = f(u) and u = g(x), then dy/dx = (dy/du) * (du/dx). Apply the chain rule when a function is composed of other functions.
Quotient Rule The quotient rule is a differentiation rule used to find the derivative of the quotient of two functions. If f(x) = u(x) / v(x), then f'(x) = (u'v - uv') / (v^2). Apply the quotient rule to functions that are ratios of two functions of x. Find the derivative of f(x) = (x^2 + 1) / x.
Definition
Formula Procedure
Example
Definition Formula
Procedure Example
Definition
Formula Procedure
Example Find the critical points of f(x) = x^3 - 6x^2 + 9x. Find the derivative: f'(x) = 3x^2 - 12x + 9 Set f'(x) = 0 to find critical points: 3x^2 - 12x + 9 = 0
Now, differentiate with respect to x: d/dx [sin^2(x) + cos^2(x)] = d/dx [1] The derivative of a constant is zero.
Implicit Differentiation Implicit differentiation is used to find the derivative of an implicitly defined function, where the dependent variable is not explicitly solved for.
dy/dx = -x / y Higher Order Derivatives Higher order derivatives are derivatives of a function taken more than once. The nth derivative of a function is denoted as f^(n)(x), where n is a positive integer. Take multiple derivatives of a function to find higher order derivatives.
as explained. Differentiate both sides of the equation with respect to x and solve for the derivative. Find dy/dx for the equation x^2 + y^2 = 1.
2y(dy/dx) = -2x dy/dx = -2x / (2y)
d/dx (x^2 + y^2) = d/dx (1)
Differentiate both sides with respect to x:
So, the derivative of sin^2(x) + cos^2(x) is zero.
f'(x) = d/dx (3x^3) = 9x^
First derivative:
Solve for dy/dx:
2x + 2y(dy/dx) = 0
Apply differentiation rules:
Second derivative: f''(x) = d/dx (9x^2) = 18x Critical Points Critical points of a function are values of x where the derivative is either zero or undefined. as explained. Find critical points by solving for x when f'(x) = 0 or when the derivative is undefined.
Find the second derivative of f(x) = 3x^3.
Factor the quadratic equation: (x - 3)(3x - 3) = 0 Solve for x: x = 3 (repeated root) The critical point is x = 3. 14 Definition
Formula Procedure
Example Find the local extrema of f(x) = x^3 - 6x^2 + 9x. Critical point found earlier: x = 3 Use the first and second derivative tests: Evaluate f'(x) at x = 3: f'(3) = 0 Evaluate f''(x) at x = 3: f''(3) = 18 (positive)
Definition
Formula Procedure
Example Find the concavity and inflection points of f(x) = x^3 - 6x^2 + 9x.
Determine concavity by analyzing the sign of the second derivative. Inflection points occur where f''(x) = 0 or changes sign.
There is an inflection point at x = 3.
x = 3 (repeated root)
(x - 3)(3x - 3) = 0
3x^2 - 12x + 9 = 0
Critical point found earlier: x = 3
To find inflection points, solve for x when f''(x) = 0:
Since f''(3) > 0, the function is concave up on the interval containing x = 3.
Evaluate f''(x) at x = 3: f''(3) = 18 (positive)
Use the second derivative to find concavity:
Concavity and Inflection Points Concavity indicates whether a function is "bending upward" (concave up) or "bending downward" (concave down) on an interval. Inflection points are points where the concavity changes. as explained.
Since f'(3) = 0 and f''(3) > 0, there is a local minimum at x = 3.
Local Extrema Local extrema are points where a function reaches a maximum (local maximum) or minimum (local minimum) value. as explained. Determine local extrema by analyzing the sign changes of the derivative around critical points.
Definition
Formula Procedure
Example
Definition
Formula
Procedure Example The limit is in the indeterminate form 0/0. Apply L'Hôpital's Rule: Limit = lim(x->0) (sin(x)/x) = lim(x->0) (cos(x)/1) = cos(0)/1 = 1/1 = 1. 20 Definition
Formula
Procedure
According to the Pythagorean theorem: x^2 + y^2 = z^2. Differentiate both sides with respect to t: 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt).
as explained. Set up an equation involving the related rates, differentiate both sides, and solve for the desired rate. Consider a right triangle where the length of one leg is increasing at a rate of 2 cm/s and the length of the other leg is decreasing at a rate of 3 cm/s. Find the rate of change of the hypotenuse when one leg is 4 cm and the other is 3 cm.
Related Rates Related rates problems involve finding the rate of change of one quantity with respect to time when it depends on another changing quantity.
Plug in the given values: 2(4)(2) + 2(3)(-3) = 2z(dz/dt).
L'Hôpital's Rule L'Hôpital's Rule is used to evaluate indeterminate forms (0/0 or ∞/∞) when finding limits If the limit of f(x)/g(x) as x approaches a is an indeterminate form, then the limit can be found by taking the derivative of both f(x) and g(x) and applying the rule.
Apply L'Hôpital's Rule by finding the limit of the ratio of derivatives.
dz/dt = -5 cm/s.
-10 = 2z(dz/dt)
8 - 18 = 2z(dz/dt)
Find the limit as x approaches 0 of (sin(x)/x)
Taylor Series and Maclaurin Series Taylor series and Maclaurin series are representations of functions as infinite series, often used for approximation. Taylor series: f(x) = Σ [f^(n)(a) / n!] * (x - a)^n, where n ranges from 0 to ∞. Maclaurin series is a Taylor series centered at a = 0. Expand a function into its Taylor or Maclaurin series using derivatives.
Solve for dz/dt:
Let x be the length of one leg (dx/dt = 2 cm/s), y be the length of the other leg (dy/dt = -3 cm/s), and z be the length of the hypotenuse (dz/dt, the rate we want
Example f(x) = e^x f'(x) = e^x f''(x) = e^x f'''(x) = e^x The Maclaurin series is given by: e^x = Σ [x^n / n!], where n ranges from 0 to ∞. This is the Maclaurin series representation of e^x. 21 Definition
Formula
Procedure Example
Definition
Formula Procedure
Example
Definition
Formula Procedure Example
Definite Integrals Definite integrals find the accumulated area under a curve between two specified limits.
∫3x^2 dx = x^3 + C, where C is the constant of integration.
Antiderivatives and Indefinite Integrals Antiderivatives are functions whose derivatives match the original function. Indefinite integrals find the family of antiderivatives. ∫f(x) dx = F(x) + C, where F(x) is the antiderivative of f(x), and C is the constant of integration. Find the antiderivative by reversing the process of differentiation Find the antiderivative of f(x) = 3x^2.
Find the Maclaurin series for f(x) = e^x.
∫[a, b] f(x) dx represents the definite integral of f(x) from a to b. Evaluate the antiderivative at the upper limit and subtract the value at the lower limit. Find ∫[1, 3] (2x + 1) dx.
Fundamental Theorem of Calculus The Fundamental Theorem of Calculus establishes a relationship between differentiation and integration. Part I relates derivatives to definite integrals, while Part II provides a method for evaluating definite integrals.
as explained.
∫[1, 3] (2x + 1) dx = [(x^2 + x) | from 1 to 3] = (3^2 + 3) - (1^2 + 1) = (9 + 3) - (1 + 1)
Apply the theorem to find derivatives or evaluate definite integrals. Use the Fundamental Theorem of Calculus to find the derivative of F(x) = ∫[0, x] (2t + 1) dt.
Definition
Formula Procedure Example
Definition Formula
Procedure
Example
Integration Techniques Integration techniques involve various methods to evaluate integrals, such as substitution, trigonometric identities, and partial fractions. as explained. Apply the appropriate technique to simplify and solve integrals. Evaluate ∫(x^2 + 2x) e^(x + 1) dx.
Continue integrating: = (x^2 + 2x)e^(x + 1) - 2(e^(x + 1) - e^(x)) + C Final result: ∫(x^2 + 2x) e^(x + 1) dx = (x^2 + 2x)e^(x + 1) - 2(e^(x + 1) - e^(x)) + C.
Apply integration by parts: ∫(x^2 + 2x) e^(x + 1) dx = (x^2 + 2x)e^(x + 1) - ∫(2x + 2)e^(x + 1) dx Now, integrate the remaining part: = (x^2 + 2x)e^(x + 1) - 2∫(x + 1)e^(x + 1) dx
Use integration by parts: Let u = (x^2 + 2x) and dv = e^(x + 1) dx. Calculate du and v: du = (2x + 2) dx v = e^(x + 1)
To find θ, use θ = arctan(y/x): θ = arctan(y/x) = arctan(0/2) = arctan(0) = 0 The polar form of the Cartesian equation is r = 2 and θ = 0.
r = √(x^2 + y^2) θ = arctan(y/x) For the equation x^2 + y^2 = 4: r = √(x^2 + y^2) = √(4) = 2
Polar Coordinates Polar coordinates describe points in a plane using a radius and an angle. Convert between polar and Cartesian coordinates using r = √(x^2 + y^2) and θ = arctan(y/x). Transform equations between polar and Cartesian coordinates when working with polar functions. Convert the Cartesian equation x^2 + y^2 = 4 into polar form. To convert to polar coordinates, use the formulas:
Definition
Formula
Procedure Example Plot the polar graph of r = 2 + 2cos(θ).
Definition
Formula Procedure
Example Find the dot product of vectors A = (2, -3, 1) and B = (4, 1, -2). The dot product of two vectors A and B is given by: A · B = (Ax * Bx) + (Ay * By) + (Az * Bz) Given vectors A = (2, -3, 1) and B = (4, 1, -2), calculate the dot product: A · B = (2 * 4) + (-3 * 1) + (1 * -2) A · B = 8 - 3 - 2 A · B = 3 The dot product of vectors A and B is 3.
When θ = π/2, r = 2 + 2cos(π/2) = 2 + 2(0) = 2
Vectors in three-dimensional space describe quantities with both magnitude and direction using components in the x, y, and z directions. Vectors are represented as (x, y, z) or in unit vector notation as ai + bj + ck. Perform vector operations, including addition, subtraction, dot product, and cross product.
Various polar functions like r = a, r = aθ, and r = a sin(θ) represent different shapes in polar graphs. Plot polar graphs by varying θ and calculating r values.
When θ = 0, r = 2 + 2cos(0) = 2 + 2(1) = 4
Vectors in Space
Plot these points and connect them to create the graph. The graph will be a cardioid, a heart-shaped curve with a loop.
When θ = 2π, r = 2 + 2cos(2π) = 2 + 2(1) = 4
When θ = 3π/2, r = 2 + 2cos(3π/2) = 2 + 2(0) = 2
To plot the polar graph of r = 2 + 2cos(θ), consider values of θ ranging from 0 to 2π.
Polar Graphs Polar graphs are representations of functions in polar coordinates, often involving symmetry and periodicity.
When θ = π, r = 2 + 2cos(π) = 2 + 2(-1) = 0
Definition
Formula Procedure
Example
x = r * sin(φ) * cos(θ) y = r * sin(φ) * sin(θ) z = r * cos(φ)
x = 1 * sin(π/3) * cos(π/4) = (√3/2) * (√2/2) = (√6/4) y = 1 * sin(π/3) * sin(π/4) = (√3/2) * (√2/2) = (√6/4) z = 1 * cos(π/3) = (1/2)
Definition
Formula Procedure
Example
Partial derivatives: ∂r/∂u = ⟨∂u/∂u, ∂v/∂u, ∂(u^2 + v^2)/∂u⟩ = ⟨1, 0, 2u⟩ ∂r/∂v = ⟨∂u/∂v, ∂v/∂v, ∂(u^2 + v^2)/∂v⟩ = ⟨0, 1, 2v⟩ Evaluate these derivatives at the point (1, 2): ∂r/∂u(1, 2) = ⟨1, 0, 4⟩ ∂r/∂v(1, 2) = ⟨0, 1, 4⟩
So, the Cartesian coordinates are (x, y, z) = ((√6/4), (√6/4), (1/2)).
Convert between spherical and Cartesian coordinates using r = √(x^2 + y^2 + z^2), θ = arctan(y/x), and φ = arccos(z/√(x^2 + y^2 + z^2)).
To convert from spherical coordinates (r, θ, φ) to Cartesian coordinates (x, y, z), use the formulas:
Convert the point (1, π/4, π/3) from spherical coordinates to Cartesian coordinates.
Spherical Coordinates Spherical coordinates describe points in three-dimensional space using a radial distance (r), an azimuthal angle (θ), and an inclination angle (φ).
r, θ, φ
Given spherical coordinates (r = 1, θ = π/4, φ = π/3), convert to Cartesian coordinates:
Parametric Surfaces Parametric surfaces represent surfaces in three-dimensional space using two parameters and vector-valued functions. r(u, v) = ⟨f(u, v), g(u, v), h(u, v)⟩ Describe surfaces using parametric equations in terms of two parameters and plot them in space. Consider the parametric surface given by r(u, v) = ⟨u, v, u^2 + v^2⟩. Find the equation of the tangent plane at the point (1, 2, 5). To find the equation of the tangent plane at the point (1, 2, 5) on the parametric surface r(u, v) = ⟨u, v, u^2 + v^2⟩, we need to find the partial derivatives and evaluate them at the point (1, 2).
N(1, 2) = ∂r/∂u(1, 2) × ∂r/∂v(1, 2) = ⟨1, 0, 4⟩ × ⟨0, 1, 4⟩ Calculate the cross product: N(1, 2) = ⟨(04 - 41), (40 - 41), (11 - 00)⟩ N(1, 2) = ⟨-4, -4, 1⟩
-4(x - 1) - 4(y - 2) + 1(z - 5) = 0 Simplify and rearrange: -4x + 4 - 4y + 8 + z - 5 = 0 -4x - 4y + z + 7 = 0 Final result:
Definition
Formula
Procedure
Example
The divergence of a vector field F = ⟨ P, Q, R ⟩ is given by:
∇ · F = ∂P/∂x + ∂Q/∂y + ∂R/∂z
Calculate the partial derivatives: ∂P/∂x = ∂(x^2)/∂x = 2x ∂Q/∂y = ∂(2y)/∂y = 2 ∂R/∂z = ∂(z^3)/∂z = 3z^ Now, sum these partial derivatives: ∇ · F = 2x + 2 + 3z^ The divergence of the vector field F is 2x + 2 + 3z^2.
Analyze and work with vector fields to understand their behavior, compute divergence, curl, and line/surface integrals. Given a vector field F(x, y, z) = ⟨x^2, 2y, z^3⟩, find the divergence ∇ · F. To find the divergence ∇ · F for the vector field F(x, y, z) = ⟨x^2, 2y, z^3⟩, we need to compute the divergence of each component and then sum them.
The equation of the tangent plane at the point (1, 2, 5) on the parametric surface is -4x - 4y + z + 7 = 0. Vector Fields Vector fields assign a vector to each point in space, representing a physical quantity with both magnitude and direction. Vector fields are often represented as F(x, y, z) = ⟨P(x, y, z), Q(x, y, z), R(x, y, z)⟩.
The equation of the tangent plane at the point (1, 2, 5) is given by:
Now, find the normal vector at the point (1, 2) by taking the cross product of the partial derivatives:
