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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 3.2.29a
Chapter 3, Problem 3.2.29a

21–30. Derivatives
a. Use limits to find the derivative function f' for the following functions f.
f(s) = 4s³+3s; a= -3, -1

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Step 1: Recall the definition of the derivative using limits. The derivative of a function f(s) at a point s is given by the limit: f'(s) = \(\lim\)_{h \(\to\) 0} \(\frac{f(s+h) - f(s)}{h}\).
Step 2: Substitute the given function f(s) = 4s^3 + 3s into the limit definition. This gives us: f'(s) = \(\lim\)_{h \(\to\) 0} \(\frac{(4(s+h)^3 + 3(s+h)) - (4s^3 + 3s)}{h}\).
Step 3: Expand the expression (s+h)^3 using the binomial theorem: (s+h)^3 = s^3 + 3s^2h + 3sh^2 + h^3. Substitute this into the limit expression.
Step 4: Simplify the expression by distributing and combining like terms. This involves expanding 4(s+h)^3 and 3(s+h), then subtracting 4s^3 + 3s.
Step 5: Factor out h from the numerator and then cancel it with the h in the denominator. Finally, take the limit as h approaches 0 to find the derivative f'(s).

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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 measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. The derivative is often denoted as f'(s) and represents the slope of the tangent line to the function's graph at a given point.
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Limit

A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value. It is used to define derivatives and integrals. In the context of finding a derivative, the limit helps to determine the instantaneous rate of change of the function at a specific point by examining the function's values as they get arbitrarily close to that point.
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Power Rule

The Power Rule is a basic differentiation rule that simplifies the process of finding the derivative of polynomial functions. It states that if f(s) = s^n, where n is a real number, then the derivative f'(s) = n * s^(n-1). This rule is particularly useful for functions like f(s) = 4s³ + 3s, allowing for quick calculation of derivatives without using the limit definition directly.
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Related Practice
Textbook Question

Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.

f(x) = 3x2 - 4x; P(1, -1)

Textbook Question

Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.

f(x) = 8 - 2x2; P(0, 8)

Textbook Question

79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>

a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)

x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)

Textbook Question

Comparing velocities Two stones are thrown vertically upward, each with an initial velocity of 48 ft/s at time t=0. One stone is thrown from the edge of a bridge that is 32 feet above the ground, and the other stone is thrown from ground level. The height above the ground of the stone thrown from the bridge after t seconds is f(t) = − 16t²+48t+32. and the height of the stone thrown from the ground after t seconds is g(t) = −16t²+48t.

a. Show that the stones reach their high points at the same time.

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).

a. For the following functions s(t), find the instantaneous velocity function v(t). (Recall that the velocity function v is the derivative of the position function s.)

s(t)= −16t²+100t

Textbook Question

Consider the following cost functions.

a. Find the average cost and marginal cost functions.

C(x) = 1000+0.1x, 0≤x≤5000, a=2000