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
Capitolo 3, Problema 3.4.19

Derivatives Find and simplify the derivative of the following functions.
f(x) = 3x⁴(2x²−1)

Guida verificata passo dopo passo
1
Identify the function as a product of two functions: f(x) = u(x) * v(x), where u(x) = 3x⁴ and v(x) = (2x²−1).
Apply the product rule for derivatives, which states that the derivative of a product u(x)v(x) is u'(x)v(x) + u(x)v'(x).
Find the derivative of u(x) = 3x⁴. The derivative, u'(x), is obtained by applying the power rule: u'(x) = 12x³.
Find the derivative of v(x) = 2x²−1. The derivative, v'(x), is obtained by applying the power rule: v'(x) = 4x.
Substitute u(x), u'(x), v(x), and v'(x) into the product rule formula: f'(x) = 12x³(2x²−1) + 3x⁴(4x). Simplify the expression to find the derivative of f(x).

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.

Derivatives

A derivative represents the rate at which a function changes at a given point. It is a fundamental concept in calculus that measures how a function's output value changes as its input value changes. The derivative can be interpreted as the slope of the tangent line to the graph of the function at a specific point.
Video consigliato:

Product Rule

The Product Rule is a formula used to find the derivative of the product of two functions. It states that if you have two functions, u(x) and v(x), the derivative of their product is given by u'v + uv'. This rule is essential when differentiating functions that are multiplied together, as in the case of f(x) = 3x⁴(2x²−1).
Video consigliato:
05:18
The Product Rule

Simplification of Derivatives

After finding the derivative of a function, simplification is often necessary to express the result in a more manageable form. This may involve combining like terms, factoring, or reducing fractions. Simplifying the derivative helps in understanding the behavior of the function and makes it easier to analyze critical points and concavity.
Video consigliato: