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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.6.76b
Chapter 3, Problem 3.6.76b

Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.


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Find the derivatives with respect to x of the following combinations at the given value of x.


b. f(x)g³(x), x = 0

Verified step by step guidance
1
Step 1: Recognize that the function to differentiate is f(x)g³(x). To find its derivative, use the product rule and the chain rule.
Step 2: Apply the product rule: If h(x) = f(x)g³(x), then h'(x) = f'(x)g³(x) + f(x) * d/dx[g³(x)].
Step 3: Use the chain rule to differentiate g³(x): d/dx[g³(x)] = 3g²(x)g'(x). Substitute this into the product rule.
Step 4: Substitute the values of f(x), g(x), f'(x), and g'(x) at x = 0 from the table into the derivative expression. Specifically, f(0) = 1, g(0) = 1, f'(0) = 5, and g'(0) = 1/3.
Step 5: Combine the terms from the product rule and chain rule to express the derivative at x = 0. Simplify the expression without calculating the final numerical value.

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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 when finding the derivative of the product of two functions. If u(x) and v(x) are differentiable functions, the derivative of their product is given by (uv)' = u'v + uv'. This rule is essential for solving problems involving the derivative of a product, such as f(x)g³(x).
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The Product Rule

Chain Rule

The chain rule is used to differentiate composite functions. If a function y = g(u) and u = f(x), then the derivative dy/dx is found by multiplying the derivative of g with respect to u by the derivative of u with respect to x, or dy/dx = (dy/du) * (du/dx). This rule is crucial when dealing with functions raised to a power, like g³(x), where g(x) is a function of x.
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Intro to the Chain Rule

Substitution of Values

Substitution involves replacing variables with specific values to evaluate expressions or derivatives at particular points. In this problem, after applying the product and chain rules, substitute x = 0 into the derived expression using the given values for f(x), g(x), f'(x), and g'(x) to find the derivative at x = 0. This step is necessary to obtain the numerical result.
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Substitution With an Extra Variable
Related Practice
Textbook Question

A sliding ladder


A 13-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft/sec.


b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?


Textbook Question

Quadratic approximations


b. Find the quadratic approximation to f(x) = 1/(1 − x) at x = 0.

Textbook Question

Consider the function f graphed here. The domain of f is the interval [−4, 6] and its graph is made of line segments joined end to end.


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b. Graph the derivative of f. The graph should show a step function.

Textbook Question

In Exercises 47 and 48, find an equation for


(b) the horizontal tangent line to the curve at Q.


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Textbook Question

The folium of Descartes (See Figure 3.27)


b. At what point other than the origin does the folium have a horizontal tangent line?


Textbook Question

Hauling in a dinghy A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow. The rope is hauled in at the rate of 2 ft/sec.


b. At what rate is the angle θ changing at this instant (see the figure)?