Textbook Question
Derivative Calculations
In Exercises 1–12, find the first and second derivatives.
y = 6x² − 10x − 5x⁻²
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 3.1.22
In Exercises 19–22, find the slope of the curve at the point indicated.
y = (x − 1) / (x + 1), x = 0
Verified step by step guidance1
To find the slope of the curve at a given point, we need to find the derivative of the function. The function given is \( y = \frac{x - 1}{x + 1} \).
Use the quotient rule to differentiate the function. The quotient rule states that if \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \). Here, \( u = x - 1 \) and \( v = x + 1 \).
Differentiate \( u \) and \( v \): \( u' = 1 \) and \( v' = 1 \).
Substitute \( u, v, u', \) and \( v' \) into the quotient rule formula: \( y' = \frac{(1)(x + 1) - (x - 1)(1)}{(x + 1)^2} \).
Simplify the expression for \( y' \) and then substitute \( x = 0 \) into the derivative to find the slope of the curve at the point \( x = 0 \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative
The derivative of a function at a point provides the slope of the tangent line to the curve at that point. It is a fundamental concept in calculus used to determine how a function changes as its input changes. For the function y = (x − 1) / (x + 1), finding the derivative will help us calculate the slope at x = 0.
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05:44
Derivatives
Quotient Rule
The quotient rule is a method for finding the derivative of a function that is the quotient of two differentiable functions. If y = u/v, where both u and v are functions of x, the derivative y' is given by (u'v - uv')/v². Applying the quotient rule to y = (x − 1) / (x + 1) will allow us to find its derivative.
Recommended video:
06:43The Quotient Rule
Substitution
Substitution involves replacing a variable with a specific value to evaluate a function or its derivative at that point. After finding the derivative of y = (x − 1) / (x + 1), substituting x = 0 into the derivative will yield the slope of the curve at the specified point.
Recommended video:
04:27Substitution With an Extra Variable
Related Practice
Textbook Question
Differential Estimates of Change
In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.
The change in the volume V = x³ of a cube when the edge lengths change from x₀ to x₀ + dx
Textbook Question
Normal lines parallel to a line Find the normal lines to the curve xy + 2x – y = 0 that are parallel to the line 2x + y = 0.
Textbook Question
Estimating height of a building A surveyor, standing 30 ft from the base of a building, measures the angle of elevation to the top of the building to be 75°. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%?
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
Graphs
Match the functions graphed in Exercises 27–30 with the derivatives graphed in the accompanying figures (a)–(d).
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