(1) A course repeater claims that his brother who did some Calculus at college years ago, currently on some cough medication, believes that the first derivative of (2x? – 1)(x² + 3) [(2x² – 1)(x² + 3) 2x + 1 4x 2х f(x) = -is f'(x) = x2 + 1 [2x² x2 + 3 x2 + x2 + 1 and (x² + 8)7 14x 30х (x² + 8)7 that the first derivative of g(x) is g'(x) = (2 – 3x²)5 [x² + 8 2 – 3x²] |(2 – 3x²)5| Using the product, qoutient and chain rules you learnt in Unit 7 of this Course, could you confirm whether his brother is right, wrong or a little tipsy.
(1) A course repeater claims that his brother who did some Calculus at college years ago, currently on some cough medication, believes that the first derivative of (2x? – 1)(x² + 3) [(2x² – 1)(x² + 3) 2x + 1 4x 2х f(x) = -is f'(x) = x2 + 1 [2x² x2 + 3 x2 + x2 + 1 and (x² + 8)7 14x 30х (x² + 8)7 that the first derivative of g(x) is g'(x) = (2 – 3x²)5 [x² + 8 2 – 3x²] |(2 – 3x²)5| Using the product, qoutient 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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![(1) A course repeater claims that his brother who did some Calculus at college years ago,
currently on some cough medication, believes that the first derivative of
(2x? – 1)(x² + 3)
[(2x² – 1)(x² + 3)
2x
+
1
4x
2х
f(x) =
-is f'(x) =
x2 + 1
[2x²
x2 + 3
x2 +
x2 + 1
and
(x² + 8)7
14x
30х
(x² + 8)7
that the first derivative of g(x)
is g'(x) =
(2 – 3x²)5
[x² + 8
2 – 3x²] |(2 – 3x²)5|
Using the product, qoutient and chain rules you learnt in Unit 7 of this Course, could you
confirm whether his brother is right, wrong or a little tipsy.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe54654cb-105a-4da5-9cf6-6527fa64119a%2Fa2d1e0fe-647c-4d02-8d91-f51b76fe6442%2Fh9nukw_processed.png&w=3840&q=75)
Transcribed Image Text:(1) A course repeater claims that his brother who did some Calculus at college years ago,
currently on some cough medication, believes that the first derivative of
(2x? – 1)(x² + 3)
[(2x² – 1)(x² + 3)
2x
+
1
4x
2х
f(x) =
-is f'(x) =
x2 + 1
[2x²
x2 + 3
x2 +
x2 + 1
and
(x² + 8)7
14x
30х
(x² + 8)7
that the first derivative of g(x)
is g'(x) =
(2 – 3x²)5
[x² + 8
2 – 3x²] |(2 – 3x²)5|
Using the product, qoutient and chain rules you learnt in Unit 7 of this Course, could you
confirm whether his brother is right, wrong or a little tipsy.
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