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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 3.91c
Chapter 3, Problem 3.91c

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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1
To solve this problem, we need to use the formula for the derivative of the inverse function. The formula is: (f^{-1})'(x) = 1 / f'(f^{-1}(x)).
First, identify f(2) from the graph of f. This value is the input for the inverse function f^{-1}.
Next, find the value of f^{-1}(f(2)) using the graph of f. This is the x-value for which f(x) equals f(2).
Once you have f^{-1}(f(2)), use the graph of f' to find f'(f^{-1}(f(2))). This is the derivative of f at the point f^{-1}(f(2)).
Finally, apply the formula: (f^{-1})'(f(2)) = 1 / f'(f^{-1}(f(2))). This will give you the derivative of the inverse function at the point f(2).

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Key Concepts

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

Inverse Function Theorem

The Inverse Function Theorem states that if a function f is continuous and differentiable, and its derivative f' is non-zero at a point, then the inverse function f^-1 exists locally around that point. This theorem is crucial for understanding how to differentiate inverse functions and relates the derivatives of f and f^-1.
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Inverse Cosine

Chain Rule

The Chain Rule is a fundamental principle in calculus that allows us to differentiate composite functions. It states that if you have a function g(f(x)), the derivative is g'(f(x)) * f'(x). This rule is essential when dealing with derivatives of inverse functions, as it helps in relating the derivatives of f and its inverse.
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Intro to the Chain Rule

Evaluating Derivatives at Specific Points

To evaluate derivatives at specific points, one must first find the value of the function at that point and then apply the appropriate derivative rules. In the context of the question, evaluating (f^-1)'(f(2)) requires finding f(2) and then using the Inverse Function Theorem to determine the derivative of the inverse function at that point.
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Related Practice
Textbook Question

{Use of Tech} Spring runoff The flow of a small stream is monitored for 90 days between May 1 and August 1. The total water that flows past a gauging station is given by v(t) = <matrix 2x2> where V is measured in cubic feet and t is measured in days, with t=0 corresponding to May 1.

c. Describe the flow of the stream over the 3-month period. Specifically, when is the flow rate a maximum?

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Textbook Question

Another way to approximate derivatives is to use the centered difference quotient: f' (a) ≈ f(a+h) - f(a- h) / 2h. Again, consider f(x) = √x.

c. Explain why it is not necessary to use negative values of h in the table of part (b).

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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

Highway travel A state patrol station is located on a straight north-south freeway. A patrol car leaves the station at 9:00 A.M. heading north with position function s = f(t) that gives its location in miles t hours after 9:00 A.M. (see figure). Assume s is positive when the car is north of the patrol station. <IMAGE>

c. Find the average velocity of the car over the interval [1.75, 2.25]. Estimate the velocity of the car at 11:00 A.M. and determine the direction in which the patrol car is moving.

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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 19.6 m/s from a height of 24.5 m above the ground. The height (in meters) of the stone above the ground t seconds after it is thrown is s(t) = -4.9t²+19.6t+24.5.

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

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Textbook Question

Vibrations of a spring Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position y=0y=0 when the mass hangs at rest. Suppose you push the mass to a position y0y_0 units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system), the position y of the mass after t seconds is y=y0cos(tkm)y=y_0\(\cos\]\left\)(t\(\sqrt{\frac{k}{m}\)}\(\right\)), where k>0k>0 is a constant measuring the stiffness of the spring (the larger the value of kk, the stiffer the spring) and yy is positive in the upward direction.

Use equation (4) to answer the following questions.

c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness (kk is increased by a factor of 44)?

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