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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.10.57

Find (f^−1)′(3), where f(x)=x³+x+1.

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Step 1: Understand the problem. We need to find the derivative of the inverse function \( (f^{-1})'(3) \) for the function \( f(x) = x^3 + x + 1 \).
Step 2: Use the formula for the derivative of an inverse function: \( (f^{-1})'(y) = \frac{1}{f'(x)} \) where \( f(x) = y \).
Step 3: Find \( x \) such that \( f(x) = 3 \). This means solving the equation \( x^3 + x + 1 = 3 \) to find the value of \( x \).
Step 4: Compute the derivative \( f'(x) \) of the function \( f(x) = x^3 + x + 1 \). The derivative is \( f'(x) = 3x^2 + 1 \).
Step 5: Evaluate \( f'(x) \) at the \( x \) found in Step 3, and use the formula from Step 2 to find \( (f^{-1})'(3) = \frac{1}{f'(x)} \).

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Key Concepts

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

Inverse Function

An inverse function reverses the effect of the original function. For a function f(x), its inverse f^(-1)(y) satisfies the equation f(f^(-1)(y)) = y. To find the derivative of an inverse function at a specific point, we often use the relationship between the derivatives of the original and inverse functions.
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Derivative of Inverse Function

The derivative of an inverse function can be calculated using the formula (f^(-1))'(y) = 1 / f'(x), where y = f(x). This means that to find the derivative of the inverse at a point, we first need to determine the corresponding x-value such that f(x) equals the given y-value, and then compute the derivative of f at that x.
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Chain Rule

The chain rule is a fundamental theorem in calculus used to differentiate composite functions. It states that if a function y = f(g(x)) is composed of two functions, the derivative is given by dy/dx = f'(g(x)) * g'(x). This rule is essential when dealing with inverse functions, as it helps in understanding how changes in x affect y through the composition of functions.
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