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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblema 4.R.107
Capitolo 4, Problema 4.R.107

104–107. Functions from derivatives Find the function f with the following properties.




h'(x) = (x⁴ -2) /(1 + x²) ; h (1) = -(2/3) 

Guida verificata passo dopo passo
1
Step 1: Recognize that the problem involves finding the original function h(x) given its derivative h'(x) = (x⁴ - 2) / (1 + x²) and an initial condition h(1) = -2/3. This is a problem of integration and using the constant of integration.
Step 2: Set up the integral to find h(x). To reverse the derivative, integrate h'(x): ∫((x⁴ - 2) / (1 + x²)) dx. This will give the general form of h(x) plus a constant of integration, C.
Step 3: Break the integral into manageable parts. Notice that the numerator (x⁴ - 2) can be split into two terms: x⁴ and -2. Rewrite the integral as ∫(x⁴ / (1 + x²)) dx - ∫(2 / (1 + x²)) dx.
Step 4: Solve each integral separately. For ∫(x⁴ / (1 + x²)) dx, use polynomial division or substitution techniques to simplify. For ∫(2 / (1 + x²)) dx, recognize it as a standard integral that results in 2 * arctan(x).
Step 5: Combine the results of the integrals and add the constant of integration, C. Use the initial condition h(1) = -2/3 to solve for C by substituting x = 1 into the expression for h(x).

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Derivatives

A derivative represents the rate of change of a function with respect to its variable. It is a fundamental concept in calculus that allows us to understand how a function behaves locally. The notation h'(x) indicates the derivative of the function h with respect to x, which can be used to find slopes of tangent lines and optimize functions.
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Derivatives

Integration

Integration is the process of finding the antiderivative of a function, essentially reversing differentiation. It allows us to recover the original function from its derivative. In this context, to find the function h from its derivative h'(x), we will need to perform integration, which may involve applying techniques such as substitution or integration by parts.
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Integration by Parts for Definite Integrals

Initial Conditions

Initial conditions are specific values that help determine a unique solution to a differential equation. In this problem, the condition h(1) = -2/3 provides a point through which the function h must pass. This information is crucial for solving the integral obtained from the derivative, as it allows us to find the constant of integration and fully define the function.
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Initial Value Problems
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