Skip to main content
Ch. 3 - Derivatives
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. 3 - DerivativesProblema 120a
Capitolo 3, Problema 120a

Let f(x) = x².
a. Show that f(x)−f(y) / x−y = f′(x+y²), for all x≠y.

Guida verificata passo dopo passo
1
Start by understanding the problem: We need to show that the difference quotient (f(x) - f(y)) / (x - y) is equal to the derivative of f evaluated at (x + y²).
First, calculate f(x) and f(y) using the given function f(x) = x². This gives us f(x) = x² and f(y) = y².
Substitute these into the difference quotient: (f(x) - f(y)) / (x - y) = (x² - y²) / (x - y).
Recognize that x² - y² is a difference of squares, which can be factored as (x - y)(x + y). Substitute this into the difference quotient to simplify: ((x - y)(x + y)) / (x - y).
Cancel the (x - y) terms in the numerator and denominator, assuming x ≠ y, to get x + y. Now, find the derivative f'(x) = 2x, and evaluate it at x + y²: f'(x + y²) = 2(x + y²). Compare this with the simplified expression x + y to verify the equality.

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:
2m

Concetti chiave

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

Difference Quotient

The difference quotient is a fundamental concept in calculus that represents the average rate of change of a function over an interval. It is defined as (f(x) - f(y)) / (x - y) for x ≠ y. This expression is crucial for understanding the derivative, as it approaches the instantaneous rate of change as y approaches x.
Video consigliato:
06:43
The Quotient Rule

Derivative

The derivative of a function at a point measures how the function's output changes as its input changes. It is denoted as f'(x) and can be interpreted as the slope of the tangent line to the function's graph at that point. In this question, we need to show that the difference quotient equals the derivative of f at a specific point, which involves applying the definition of the derivative.
Video consigliato:
Percorso guidato
05:44
Derivatives

Function Composition

Function composition involves combining two functions to create a new function, where the output of one function becomes the input of another. In this context, we need to evaluate f' at (x + y²), which requires understanding how to apply the derivative to a function that is itself a composition of variables. This concept is essential for manipulating and simplifying expressions involving derivatives.
Video consigliato:
3:48
Evaluate Composite Functions - Special Cases
Pratica correlata
Domanda del libro di testo

Water flows into a conical tank at a rate of 2 ft³/min. If the radius of the top of the tank is 4 ft and the height is 6 ft, determine how quickly the water level is rising when the water is 2 ft deep in the tank.

286
views
Domanda del libro di testo

A general proof of the Chain Rule Let f and g be differentiable functions with h(x)=f(g(x)). For a given constant a, let u=g(a) and v=g(x), and define H (v) = <1x1 matrix>

c. Show that h′(a) = lim x→a ((H(g(x))+f′(g(a)))⋅g(x)−g(a)/x−a).

415
views
Domanda del libro di testo

Derivatives of even and odd functions Recall that f is even if f(−x) = f(x), for all x in the domain of f, and f is odd if f(−x) = −f(x) for all x in the domain of f.

b. If f is a differentiable, odd function on its domain, determine whether f' is even, odd, or neither.

503
views