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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 3.2.29b
Chapter 3, Problem 3.2.29b

21–30. Derivatives
b. Evaluate f'(a) for the given values of a.
f(s) = 4s³+3s; a= -3, -1

Verified step by step guidance
1
Step 1: Identify the function f(s) = 4s^3 + 3s and the values of a for which we need to evaluate the derivative, which are a = -3 and a = -1.
Step 2: Find the derivative of the function f(s) with respect to s. Use the power rule, which states that the derivative of s^n is n*s^(n-1).
Step 3: Apply the power rule to each term in the function. The derivative of 4s^3 is 12s^2, and the derivative of 3s is 3.
Step 4: Combine the derivatives to get the derivative function f'(s) = 12s^2 + 3.
Step 5: Evaluate the derivative at the given values of a. Substitute s = -3 into f'(s) to find f'(-3), and substitute s = -1 into f'(s) to find f'(-1).

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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 any given point. The derivative can be computed using various rules, such as the power rule, product rule, and quotient rule.
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Derivatives

Power Rule

The power rule is a basic differentiation technique used to find the derivative of functions in the form f(s) = s^n, where n is a real number. According to this rule, the derivative f'(s) is calculated as n * s^(n-1). This rule simplifies the process of differentiation for polynomial functions, making it easier to evaluate derivatives at specific points.
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Evaluating Derivatives at Specific Points

Evaluating the derivative at specific points involves substituting the given values into the derivative function. This process allows us to determine the instantaneous rate of change of the original function at those points. In the context of the question, we will first find the derivative of f(s) and then substitute a = -3 and a = -1 to find f'(-3) and f'(-1).
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Related Practice
Textbook Question

Derivatives of inverse functions from a table Use the following tables to determine the indicated derivatives or state that the derivative cannot be determined. <IMAGE>

b. (f^-1)'(6)

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Velocity of a car The graph shows the position s=f(t) of a car t hours after 5:00 P.M. relative to its starting point s=0,where s is measured in miles. <IMAGE>

b. At approximately what time is the car traveling the fastest? The slowest?

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Textbook Question

31–32. Velocity functions A projectile is fired vertically upward into the air, and its position (in feet) above the ground after t seconds is given by the function s(t).

b. Determine the instantaneous velocity of the projectile at t=1 and t = 2 seconds.

s(t)= −16t²+100t

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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)

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For what values of x does g(x) = x−sin x have a slope of 1?

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109-112 {Use of Tech} Calculating limits The following limits are the derivatives of a composite function g at a point a.

b. Use the Chain Rule to find each limit. Verify your answer by using a calculator.

limx→2(x2−3)5−1x−2{\(\displaystyle\]\lim\)_{x\(\to\)2}}\(\frac{\left(x^2-3\right)^5-1}{x-2}\)

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