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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))

Verified step by step guidance
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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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} 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?

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?

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).

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)=(sec^−1 x)/x on [1,∞)

Textbook Question

Derivatives of sin^n x Calculate the following derivatives using the Product Rule.

c. d/dx (sin⁴ x)

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?