Textbook Question
In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.
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x + √xy = 6, (4, 1)
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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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 3.1.43
[Technology Exercise]
Graph the curves in Exercises 39–48.
a. Where do the graphs appear to have vertical tangent lines?
b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38.
y = 4x²/⁵ − 2x
Verified step by step guidance1
Step 1: Begin by understanding the concept of a vertical tangent line. A vertical tangent occurs where the derivative of the function is undefined or infinite. For the function y = 4x^(2/5) - 2x, we need to find its derivative.
Step 2: Differentiate the function y = 4x^(2/5) - 2x with respect to x. Use the power rule for differentiation: if y = x^n, then dy/dx = n*x^(n-1). Apply this to each term in the function.
Step 3: The derivative of y = 4x^(2/5) is (2/5)*4*x^((2/5)-1) = (8/5)*x^(-3/5). The derivative of -2x is simply -2. Therefore, the derivative of the function is dy/dx = (8/5)*x^(-3/5) - 2.
Step 4: Identify where the derivative dy/dx = (8/5)*x^(-3/5) - 2 is undefined or infinite. This occurs when x^(-3/5) is undefined, which happens when x = 0. Thus, x = 0 is a candidate for a vertical tangent line.
Step 5: Confirm the vertical tangent by calculating the limit of the derivative as x approaches 0. Evaluate the limit of (8/5)*x^(-3/5) - 2 as x approaches 0 from the left and right. If the limit approaches infinity or negative infinity, a vertical tangent exists at x = 0.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Vertical Tangent Lines
Vertical tangent lines occur at points on a curve where the derivative is undefined or infinite. This typically happens when the slope of the tangent line becomes vertical, indicating a sharp turn or cusp in the graph. Identifying these points requires analyzing the behavior of the derivative as it approaches certain values.
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Slopes of Tangent Lines
Limit Calculations
Limit calculations are used to determine the behavior of a function as it approaches a specific point. In the context of vertical tangents, limits help confirm where the derivative becomes infinite or undefined. Calculating limits involves evaluating the function's behavior as the input approaches the point of interest, often using algebraic manipulation or L'Hôpital's Rule.
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05:50One-Sided Limits
Derivative of a Function
The derivative of a function represents the rate of change or slope of the function at any given point. For the function y = 4x²/⁵ − 2x, finding the derivative involves applying differentiation rules, such as the power rule, to each term. The derivative is crucial for identifying tangent lines and understanding the function's behavior, especially in determining where vertical tangents may occur.
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06:30Derivatives of Other Trig Functions
Related Practice
Textbook Question
Suppose that the function v in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say? What does this say about the Derivative Constant Multiple Rule?
Textbook Question
In Exercises 53 and 54, find both dy/dx (treating y as a differentiable function of x) and dx/dy (treating x as a differentiable function of y). How do dy/dx and dx/dy seem to be related?
54. x³ + y² = sin²y
Textbook Question
Falling meteorite The velocity of a heavy meteorite entering Earth’s atmosphere is inversely proportional to √s when it is s km from Earth’s center. Show that the meteorite’s acceleration is inversely proportional to s².
Textbook Question
One-Sided Derivatives
Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P.
Textbook Question
Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is,
d/dx (x⁻ᵐ) = −mx⁻ᵐ⁻¹
where m is a positive integer.