Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problema 34a

Capitolo 4, Problema 34a

Finding Functions from Derivatives

In Exercises 31–36, find all possible functions with the given derivative.

a. y' = 1 / 2√x

Guida verificata passo dopo passo

Step 1: Recognize that finding a function from its derivative involves integration. The given derivative is \( y' = \frac{1}{2\sqrt{x}} \).

Step 2: Rewrite the derivative in a form that is easier to integrate. Note that \( \frac{1}{2\sqrt{x}} \) can be expressed as \( \frac{1}{2}x^{-1/2} \).

Step 3: Integrate the expression \( \frac{1}{2}x^{-1/2} \) with respect to \( x \). Use the power rule for integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n \neq -1 \).

Step 4: Apply the power rule to \( \frac{1}{2}x^{-1/2} \). The integral becomes \( \frac{1}{2} \int x^{-1/2} \, dx = \frac{1}{2} \cdot \frac{x^{1/2}}{1/2} + C \). Simplify this expression.

Step 5: Simplify the result from the integration to find the general form of the function \( y(x) \). Don't forget to include the constant of integration \( C \), which represents all possible functions that have the given derivative.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Antiderivatives

Antiderivatives, or indefinite integrals, are functions that reverse the process of differentiation. To find a function from its derivative, you need to determine its antiderivative. This involves integrating the derivative function, which in this case is y' = 1 / 2√x, to find the original function y(x).

Video consigliato:

Percorso guidato

05:50

Antiderivatives

Integration Techniques

Integration techniques are methods used to find antiderivatives. For the derivative y' = 1 / 2√x, recognizing the form of the function is crucial. The expression 1 / 2√x can be rewritten as x^(-1/2), which is a power function. Applying the power rule for integration, you add 1 to the exponent and divide by the new exponent to find the antiderivative.

Video consigliato:

Percorso guidato

05:04

Introduction to Indefinite Integrals

Constant of Integration

The constant of integration is an arbitrary constant added to the antiderivative of a function. When integrating a derivative to find the original function, the constant accounts for any vertical shifts in the graph of the function. Since differentiation eliminates constants, the antiderivative must include this constant to represent all possible original functions.

Video consigliato:

Percorso guidato

05:04

Introduction to Indefinite Integrals

Pratica correlata

Domanda del libro di testo

Graphs and Graphing

Graph the curves in Exercises 33–42.

y = 𝓍³ (8―𝓍 )

185

views

Domanda del libro di testo

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

33. y = (x² - x + 1) / (x - 1)

155

views

Domanda del libro di testo

Finding Functions from Derivatives

In Exercises 31–36, find all possible functions with the given derivative.

b. y′ = x²

204

views

Domanda del libro di testo

Finding Functions from Derivatives

In Exercises 31–36, find all possible functions with the given derivative.