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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
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Capitolo 3, Problema 22

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

Guida verificata passo dopo passo
1
Identify the composite function structure: The given function is y = e^{\(\sqrt{x}\)}. This can be seen as a composition of two functions.
Choose the inner function: Let u = g(x) = \(\sqrt{x}\). This represents the inner function, which is the square root of x.
Choose the outer function: Let y = f(u) = e^u. This represents the outer function, which is the exponential function with base e.
Differentiate the inner function: Find the derivative of u with respect to x, which is \(\frac{du}{dx}\) = \(\frac{1}{2\sqrt{x}\)}.
Differentiate the outer function: Find the derivative of y with respect to u, which is \(\frac{dy}{du}\) = e^u. Then, use the chain rule to find \(\frac{dy}{dx}\) = \(\frac{dy}{du}\) \(\cdot\) \(\frac{du}{dx}\).

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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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Evaluate Composite Functions - Special Cases

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

Exponential Functions

Exponential functions are mathematical functions of the form y = a^x, where 'a' is a constant and 'x' is the variable. In the given question, the function involves the exponential function e raised to the power of another function, √x. Understanding the properties of exponential functions, including their derivatives, is vital for accurately calculating dy/dx in this context.
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Exponential Functions
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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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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

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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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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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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.

a. Graph the position function.

f(t)=6t3+36t2−54t;0≤t≤4f(t)=6t^3+36t^2-54t;0\(\le\) t\(\le\)4

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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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