Textbook Question
Derivatives and tangent lines
a. For the following functions and values of a, find f′(a).
f(x) = 8x; a = −3
227
views
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.2.25b
21–30. Derivatives
b. Evaluate f'(a) for the given values of a.
f(x) = 1/x+1; a = -1/2;5
Verified step by step guidance1
Step 1: Identify the function f(x) = \(\frac{1}{x}\) + 1 and the points a = -\(\frac{1}{2}\) and a = 5 where we need to evaluate the derivative f'(a).
Step 2: Find the derivative of the function f(x). The function can be rewritten as f(x) = x^{-1} + 1. Use the power rule for derivatives, which states that \(\frac{d}{dx}\)[x^n] = nx^{n-1}, to find f'(x).
Step 3: Apply the power rule to f(x) = x^{-1} + 1. The derivative of x^{-1} is -x^{-2}, and the derivative of a constant (1) is 0. Therefore, f'(x) = -x^{-2}.
Step 4: Evaluate f'(x) at a = -\(\frac{1}{2}\). Substitute x = -\(\frac{1}{2}\) into f'(x) = -x^{-2} to find f'(-\(\frac{1}{2}\)).
Step 5: Evaluate f'(x) at a = 5. Substitute x = 5 into f'(x) = -x^{-2} to find f'(5).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
7mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivatives
A derivative represents the rate of change of a function with respect to its variable. It is a fundamental concept in calculus that provides information about the slope of the tangent line to the curve of the function at a given point. The derivative can be computed using various rules, such as the power rule, product rule, and quotient rule, depending on the form of the function.
Recommended video:
05:44
Derivatives
Quotient Rule
The quotient rule is a method for finding the derivative of a function that is the ratio of two other functions. If f(x) = g(x)/h(x), the derivative f'(x) is given by f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2. This rule is essential when dealing with functions that involve division, such as the function f(x) = 1/(x+1) in the given question.
Recommended video:
06:43The Quotient Rule
Evaluating Derivatives at Specific Points
Evaluating the derivative at a specific point involves substituting the value of that point into the derivative function. This process provides the slope of the tangent line to the function at that particular point, which can be useful for understanding the behavior of the function. In this case, we need to compute f'(-1/2) and f'(5) to analyze the function's behavior at these values.
Recommended video:
04:50Critical Points
Related Practice
Textbook Question
45–50. Tangent lines Carry out the following steps. <IMAGE>
b. Determine an equation of the line tangent to the curve at the given point.
x³+y³=2xy; (1, 1)
269
views
Textbook Question
Use definition (1) (p. 133) to find the slope of the line tangent to the graph of f at P.
f(x) = -3x2 - 5x + 1; P(1,-7)
218
views
Textbook Question
Derivatives using tables Let and . Use the table to compute the following derivatives.
<IMAGE>
a.
284
views
Textbook Question
{Use of Tech} Spring oscillations A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance x (in inches) of the mass from its equilibrium position after t seconds is given by the function x(t) = 10sin t - 10cos t, where x is positive when the mass is above the equilibrium position. <IMAGE>
b. Find dx/dt and interpret the meaning of this derivative.
208
views
Textbook Question
45–50. Tangent lines Carry out the following steps. <IMAGE>
b. Determine an equation of the line tangent to the curve at the given point.
x⁴-x²y+y⁴=1; (−1, 1)
220
views