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2x* + х 'f(x) : %3D .3 x' - 1 2х - 3 y = 2 (x² + 4)² b. 2x – 1 y = С. Зх + 5

2x* + х 'f(x) : %3D .3 x' - 1 2х - 3 y = 2 (x² + 4)² b. 2x – 1 y = С. Зх + 5

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question

Please solve the derivative of the equation below using the derivative theorem!

2x? + x а. f(x) x³ – 1 3 2х — 3 y = (x² + 4)? b. 2x? y = Зх + 5 1 С.
Transcribed Image Text:2x? + x а. f(x) x³ – 1 3 2х — 3 y = (x² + 4)? b. 2x? y = Зх + 5 1 С.
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