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Problem Set D (1) A course repeater claims that his brother who did some Calculus at college years ago, but currently on some cough medication, believes that the first derivative of (2x2 – 1)(x2 + 3) 4x 2х 2х [(2x² – 1)(x² + 3) f(x) -is f'(x) = x2 + 1 [2x² – 1 x2 + 3 x2 + 1 x2 + 1 and (x2 + 8)7 14x 30x (x² + 8)7 that the first derivative of g(x) : (2 – 3x2)5 is g'(x) x2 + 8*2- 3x1 2– 3x²)5 2 - 3x2 - 3x2)5] Using the product, goutient and chain rules you learnt in Unit 7 of this Course, could you confirm whether his brother is right, wrong or a little tipsy.

Problem Set D (1) A course repeater claims that his brother who did some Calculus at college years ago, but currently on some cough medication, believes that the first derivative of (2x2 – 1)(x2 + 3) 4x 2х 2х [(2x² – 1)(x² + 3) f(x) -is f'(x) = x2 + 1 [2x² – 1 x2 + 3 x2 + 1 x2 + 1 and (x2 + 8)7 14x 30x (x² + 8)7 that the first derivative of g(x) : (2 – 3x2)5 is g'(x) x2 + 8*2- 3x1 2– 3x²)5 2 - 3x2 - 3x2)5] Using the product, goutient and chain rules you learnt in Unit 7 of this Course, could you confirm whether his brother is right, wrong or a little tipsy.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Problem Set D (1) A course repeater claims that his brother who did some Calculus at college years ago, but currently on some cough medication, believes that the first derivative of (2x² – 1)(x² + 3) f(x) = 4х 2x 2х (2x² – 1)(x² + 3) - is f'(x) = | %3D x² + 1 2x2 - 1 x2 +3 x2 + 1. x2 + 1 and (x² + 8)7 (x² + 8)7 (2-3x²)5 Using the product, qoutient and chain rules you learnt in Unit 7 of this Course, could you 14x 30x that the first derivative of g(x) is g'(x) = %3D [x² + 8 2 (2 – 3x²)5 2 - 3x2. confirm whether his brother is right, wrong or a little tipsy.
Transcribed Image Text:Problem Set D (1) A course repeater claims that his brother who did some Calculus at college years ago, but currently on some cough medication, believes that the first derivative of (2x² – 1)(x² + 3) f(x) = 4х 2x 2х (2x² – 1)(x² + 3) - is f'(x) = | %3D x² + 1 2x2 - 1 x2 +3 x2 + 1. x2 + 1 and (x² + 8)7 (x² + 8)7 (2-3x²)5 Using the product, qoutient and chain rules you learnt in Unit 7 of this Course, could you 14x 30x that the first derivative of g(x) is g'(x) = %3D [x² + 8 2 (2 – 3x²)5 2 - 3x2. confirm whether his brother is right, wrong or a little tipsy.
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