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 8a

Find the derivative the following ways:
Using the Product Rule or the Quotient Rule. Simplify your result.
g(t) = (t + 1)(t² - t + 1)

Guida verificata passo dopo passo
1
Identify the function as a product of two functions: \( u(t) = t + 1 \) and \( v(t) = t^2 - t + 1 \).
Recall the Product Rule for derivatives, which states: \( (uv)' = u'v + uv' \).
Find the derivative of \( u(t) = t + 1 \), which is \( u'(t) = 1 \).
Find the derivative of \( v(t) = t^2 - t + 1 \), which is \( v'(t) = 2t - 1 \).
Apply the Product Rule: \( g'(t) = u'(t)v(t) + u(t)v'(t) = 1(t^2 - t + 1) + (t + 1)(2t - 1) \). Simplify the expression to get the final 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:
3m

Concetti chiave

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

Product Rule

The Product Rule is a formula used to find the derivative of the product of two functions. If you have two functions, u(t) and v(t), the derivative of their product is given by u'v + uv'. This rule is essential when differentiating expressions where two functions are multiplied together, as it allows for a systematic approach to finding the derivative.
Video consigliato:
05:18
The Product Rule

Quotient Rule

The Quotient Rule is used to differentiate a function that is the quotient of two other functions. If f(t) = u(t)/v(t), the derivative is given by (u'v - uv')/v². This rule is particularly useful when dealing with fractions of functions, ensuring that the differentiation accounts for both the numerator and the denominator.
Video consigliato:
06:43
The Quotient Rule

Simplification of Derivatives

After applying the Product or Quotient Rule, the resulting derivative often needs simplification to make it more manageable or interpretable. This involves combining like terms, factoring, or reducing fractions. Simplification is crucial for clearer analysis and understanding of the behavior of the function, especially when evaluating limits or finding critical points.
Video consigliato: