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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 23
Chapter 3, Problem 23

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 = tan 5x²

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1
Identify the composite function structure: The given function is y = \(\tan\)(5x^2). This can be expressed in the form y = f(g(x)), where g(x) is the inner function and f(u) is the outer function.
Define the inner function: Let u = g(x) = 5x^2. This represents the expression inside the tangent function.
Define the outer function: The outer function is y = f(u) = \(\tan\)(u). This represents the tangent of the inner function.
Differentiate the inner function: Find the derivative of u with respect to x, which is \(\frac{du}{dx}\) = \(\frac{d}{dx}\)(5x^2) = 10x.
Differentiate the outer function: Find the derivative of y with respect to u, which is \(\frac{dy}{du}\) = \(\sec\)^2(u).
Apply the chain rule: Use the chain rule to find \(\frac{dy}{dx}\) = \(\frac{dy}{du}\) \(\cdot\) \(\frac{du}{dx}\). Substitute the derivatives found in the previous steps: \(\frac{dy}{dx}\) = \(\sec\)^2(5x^2) \(\cdot\) 10x.

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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 express the function y = tan(5x²) as y = f(g(x)), where g(x) is the inner function and f(u) is the outer function. Understanding how to identify these functions is crucial for applying the chain rule in differentiation.
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Evaluate Composite Functions - Special Cases

Chain Rule

The chain rule is a fundamental theorem 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, which is essential for solving the given problem.
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Intro to the Chain Rule

Inner and Outer Functions

In a composite function, the inner function is the one that is applied first, while the outer function is applied to the result of the inner function. For the function y = tan(5x²), the inner function can be identified as g(x) = 5x² and the outer function as f(u) = tan(u). Recognizing these functions is key to correctly applying the chain rule and finding the derivative.
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Related Practice
Textbook Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t2 + 32t + 48.

Determine the velocity v of the stone after t seconds.

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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 = e^4x²+1

217
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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 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t2 + 32t + 48.

With what velocity does the stone strike the ground?

240
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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 = e^√x

270
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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) = 6t3 + 36t2 - 54t; 0 ≤ t ≤ 4

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Textbook Question

Throwing a stone Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t)=−16t2+32t+48s(t)=-16t^2+32t+48 .

d. When does the stone strike the ground?

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