Textbook Question
For Exercises 55 and 56, evaluate each limit by first converting each to a derivative at a particular x-value.
lim (x → 1) (x⁵⁰ − 1) / (x − 1)
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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.
Recommended video:
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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
Derivative Calculations
In Exercises 1–12, find the first and second derivatives.
y = 6x² − 10x − 5x⁻²
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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
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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.
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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%?
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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.
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