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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.2.25b
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 guidance
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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).

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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.
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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.
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The 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.
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Critical Points
Related Practice
Textbook Question

A race Jean and Juan run a one-lap race on a circular track. Their angular positions on the track during the race are given by the functions θ(t) and ϕ(t), respectively, where 0≤t≤4 and t is measured in minutes (see figure). These angles are measured in radians, where θ=ϕ=0 represent the starting position and θ=ϕ=2π represent the finish position. The angular velocities of the runners are θ′(t) and ϕ′(t). <IMAGE>

b. Which runner has the greater average angular velocity?

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Textbook Question

Derivatives using tables Let h(x)=f(g(x))h(x)=f(g(x)) and p(x)=g(f(x))p(x)=g(f(x)). Use the table to compute the following derivatives.

<IMAGE>

b. h′(2)h^{\(\prime\)}\(\left\)(2\(\right\))

Textbook Question

Derivatives and tangent lines

b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.

f(x) = √3x; a= 12

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Textbook Question

13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

x = e^y; (2, ln 2)

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Textbook Question

13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

cos y = x; (0, π/2)

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.  

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