Textbook Question
In Exercises 5–10, find an equation for the tangent line to the curve at the given point. Then sketch the curve and tangent line together.
y = (1 / x²), (−1, 1)
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.6.4
Derivative Calculations
In Exercises 1–8, given y = f(u) and u = g(x), find dy/dx = f'(g(x)) g'(x).
y = cos u, u = −x/3
Verified step by step guidance1
First, identify the functions involved: y = cos(u) and u = -x/3. We need to find dy/dx using the chain rule.
The chain rule states that dy/dx = f'(u) * g'(x), where f(u) = cos(u) and u = g(x) = -x/3.
Calculate f'(u). Since y = cos(u), the derivative f'(u) with respect to u is -sin(u).
Next, calculate g'(x). Since u = -x/3, the derivative g'(x) with respect to x is -1/3.
Combine the derivatives using the chain rule: dy/dx = f'(u) * g'(x) = (-sin(u)) * (-1/3). Substitute u = -x/3 into the expression to get dy/dx = (1/3)sin(-x/3).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Chain Rule
The chain rule is a fundamental differentiation technique used when dealing with composite functions. It states that if you have a function y = f(u) and u = g(x), then the derivative dy/dx is found by multiplying the derivative of the outer function f'(u) by the derivative of the inner function g'(x). This allows us to differentiate complex functions by breaking them down into simpler parts.
Recommended video:
05:02
Intro to the Chain Rule
Derivative of Trigonometric Functions
Understanding the derivatives of trigonometric functions is crucial for solving calculus problems involving these functions. For example, the derivative of cos(u) with respect to u is -sin(u). This knowledge is essential when applying the chain rule to differentiate composite functions involving trigonometric expressions.
Recommended video:
6:04Introduction to Trigonometric Functions
Differentiation of Linear Functions
Differentiating linear functions is straightforward, as the derivative of a linear function ax + b is simply the constant a. In the given problem, u = -x/3 is a linear function, and its derivative with respect to x is -1/3. This derivative is used in conjunction with the chain rule to find dy/dx for the composite function y = cos(u).
Recommended video:
07:17Linearization
Related Practice
Textbook Question
In Exercises 41–44, determine whether the piecewise-defined function is differentiable at x = 0.
f(x) = { 2x − 1, x ≥ 0
x² + 2x + 7, x < 0
Textbook Question
Find the derivatives of the functions in Exercises 1–42.
𝔂 = x⁻¹/² sec (2x)²
Textbook Question
Second Derivatives
Find y'' in Exercises 59–64.
y = x(2x + 1)⁴
Textbook Question
Find the derivatives of the functions in Exercises 1–42.
𝔂 = (x + 1)² (x² + 2x)
Textbook Question
The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is
A(q) = (km / q) + cm + (hq / 2),
where q is the quantity you order when things run low (shoes, TVs, brooms, or whatever the item might be); k is the cost of placing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security).
Find dA/dq and d²A/dq².