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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 3.85c
Chapter 3, Problem 3.85c

Finding derivatives from a table Find the values of the following derivatives using the table. <IMAGE>


c. d/dx ((f(x)g(x)) |x=3

Verified step by step guidance
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Step 1: Recall the product rule for derivatives, which states that if you have two functions f(x) and g(x), the derivative of their product is given by (f(x)g(x))' = f'(x)g(x) + f(x)g'(x).
Step 2: Identify the values you need from the table. You will need f(3), g(3), f'(3), and g'(3) to apply the product rule at x = 3.
Step 3: Substitute the values from the table into the product rule formula. This means replacing f(x) with f(3), g(x) with g(3), f'(x) with f'(3), and g'(x) with g'(3).
Step 4: Calculate each term separately. First, compute f'(3)g(3) and then f(3)g'(3).
Step 5: Add the results from Step 4 to find the derivative of the product at x = 3.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Product Rule

The Product Rule is a fundamental differentiation rule used to find the derivative of the product of two functions. It states that if you have two functions, f(x) and g(x), the derivative of their product is given by d/dx(f(x)g(x)) = f'(x)g(x) + f(x)g'(x). This rule is essential for solving problems involving the multiplication of functions.
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The Product Rule

Evaluating Derivatives at a Point

Evaluating derivatives at a specific point involves substituting the value of x into the derivative expression after it has been calculated. In this case, after applying the Product Rule, you will substitute x = 3 into the resulting expression to find the specific value of the derivative at that point. This step is crucial for obtaining numerical answers from derivative expressions.
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Using Derivative Tables

Derivative tables provide pre-calculated values of derivatives for various functions at specific points. When solving derivative problems, such as the one presented, these tables can be used to quickly find f'(3) and g'(3) without needing to compute the derivatives from scratch. This efficiency is particularly useful in problems where time or complexity is a factor.
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Related Practice
Textbook Question

{Use of Tech} Power and energy Power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up. It is measured in units of joules or Calories, where 1 Cal=4184 J. One hour of walking consumes roughly 10⁶J, or 240 Cal. On the other hand, power is the rate at which energy is used, which is measured in watts, where 1 W=1 J/s. Other useful units of power are kilowatts (1 kW=10³ W) and megawatts (1 MW=10⁶ W). If energy is used at a rate of 1 kW for one hour, the total amount of energy used is 1 kilowatt-hour (1 kWh=3.6×10⁶ J) Suppose the cumulative energy used in a large building over a 24-hr period is given by E(t)=100t+4t²−t³ / 9kWh where t=0 corresponds to midnight.

c. Graph the power function and interpret the graph. What are the units of power in this case?

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Textbook Question

62–65. {Use of Tech} Graphing f and f'

c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.

f(x)=(x²−1)sin^−1 x on [−1,1]

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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>

c. At what times is the velocity of the mass zero?

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Textbook Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 64 ft/s from a height of 32 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t²+64t+32.

c. What is the height of the stone at the highest point?

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Textbook Question

Derivatives from a graph If possible, evaluate the following derivatives using the graphs of f and f'. <IMAGE>

c. (f^-1)'(f(2))

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Textbook Question

Witch of Agnesi Let y(x²+4)=8 (see figure). <IMAGE>

c. Solve the equation y(x²+4)=8 for y to find an explicit expression for y and then calculate dy/dx.

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