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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 16
Chapter 3, Problem 16

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 = (5x²+11x)^4/3

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Identify the inner function and the outer function: Let the inner function be \( u = g(x) = 5x^2 + 11x \) and the outer function be \( y = f(u) = u^{4/3} \).
Differentiate the inner function \( u = g(x) \) with respect to \( x \): \( \frac{du}{dx} = \frac{d}{dx}(5x^2 + 11x) \).
Differentiate the outer function \( y = f(u) \) with respect to \( u \): \( \frac{dy}{du} = \frac{d}{du}(u^{4/3}) \).
Apply the chain rule to find \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
Substitute the expressions for \( \frac{dy}{du} \) and \( \frac{du}{dx} \) into the chain rule formula to express \( \frac{dy}{dx} \) in terms of \( x \).

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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 given problem, 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 problem.
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Intro to the Chain Rule

Power Rule

The power rule is a basic differentiation rule that states if y = x^n, then dy/dx = n*x^(n-1). In the context of the given function, which involves a power of a polynomial, applying the power rule to the outer function will be necessary to compute the derivative correctly. This rule simplifies the process of differentiation for polynomial expressions.
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Related Practice
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 acceleration of the object when its velocity is zero.

f(t) = 2t2 - 9t + 12; 0 ≤ t ≤ 3

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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) = t2 − 4t; 0 ≤ t ≤ 5

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 = sin⁵x

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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) = t2 - 4t; 0 ≤ t ≤ 5

Textbook Question

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

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

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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 acceleration of the object when its velocity is zero.

f(t) = t2 - 4t; 0 ≤ t ≤ 5