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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 19

5–24. For each of the following composite functions, find an inner function u=g(x) and an outer function y=f(u) such that y=f(g(x)). Then calculate dy/dx.
y = √x²+1

Verified step by step guidance
1
Identify the composite function structure: The given function is y = \(\sqrt{x^2 + 1}\). This can be seen as a composition of two functions.
Define the inner function: Let u = g(x) = x^2 + 1. This is the expression inside the square root.
Define the outer function: Let y = f(u) = \(\sqrt{u}\). This represents the square root function applied to u.
Differentiate the outer function with respect to u: The derivative of y = \(\sqrt{u}\) with respect to u is dy/du = \(\frac{1}{2\sqrt{u}\)}.
Differentiate the inner function with respect to x: The derivative of u = x^2 + 1 with respect to x is du/dx = 2x.
Apply the chain rule to find dy/dx: Use the chain rule, which states that dy/dx = (dy/du) * (du/dx). Substitute the derivatives found in the previous steps.

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Key Concepts

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

Composite Functions

A composite function is formed when one function is applied to the result of another function. In the context of the question, we need to identify an inner function g(x) and an outer function f(u) such that the overall function can be expressed as y = f(g(x)). Understanding how to decompose a function into its components is essential for differentiation.
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Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that if y = f(g(x)), then the derivative dy/dx can be calculated as dy/dx = f'(g(x)) * g'(x). This rule allows us to find the derivative of complex functions by breaking them down into simpler parts, making it crucial for solving the given problem.
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Intro to the Chain Rule

Differentiation of Square Roots

Differentiating functions that involve square roots requires understanding how to apply the power rule and the chain rule effectively. For example, the derivative of √u is (1/2)u^(-1/2) * du/dx. In the given function y = √(x² + 1), recognizing how to handle the square root and the inner function is key to finding the correct derivative.
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Related Practice
Textbook Question

5–24. For each of the following composite functions, find an inner function u=g(x) and an outer function y=f(u) such that y=f(g(x)). Then calculate dy/dx.

y = √7x-1

Textbook Question

Use definition (1) (p. 133) to find the slope of the line tangent to the graph of f at P.

f(x) = √(3x + 3); P(2,3)

Textbook Question

Use definition (1) (p. 133) to find the slope of the line tangent to the graph of f at P.

f(x) = 4/x2; P(-1,4)

Textbook Question

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.

On what intervals is the speed increasing?

f(t) = 18t - 3t2; 0 ≤ t ≤ 8

Textbook Question

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.

Determine the velocity and acceleration of the object at t = 1. 

f(t) = 2t3 - 21t2 + 60t; 0 ≤ t ≤ 6

Textbook Question

Position, velocity, and acceleration Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.

b. Find and graph the velocity function. When is the object stationary, moving to the right, and moving to the left?

f(t) = 18t-3t²; 0 ≤ t ≤ 8