(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 (2x? – 1)(x² + 3) 4x 2х 2х 1[(2x2 — 1)(x? + 3) f(x) = is f'(x) = x2 + 1 |2x2 — 1 х2 + 3 х2 + 1] x2 + 1 and (x² + 8)7 (2 — Зx2)5| Using the product, qoutient and chain rules you learnt in Unit 7 of this Course, could you (x² + 8)7 30х + Lx² + 8 ' 2 – 3x². 14x that the first derivative of g(x) is g'(x) = (2 — Зх2)5 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, but currently on some cough medication, believes that the first derivative of (2x? – 1)(x² + 3) 4x 2х 2х 1[(2x2 — 1)(x? + 3) f(x) = is f'(x) = x2 + 1 |2x2 — 1 х2 + 3 х2 + 1] x2 + 1 and (x² + 8)7 (2 — Зx2)5| Using the product, qoutient and chain rules you learnt in Unit 7 of this Course, could you (x² + 8)7 30х + Lx² + 8 ' 2 – 3x². 14x that the first derivative of g(x) is g'(x) = (2 — Зх2)5 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, but
currently on some cough medication, believes that the first derivative of
(2x² – 1)(x² + 3)
4x
[(2x² – 1)(x² + 3)
2x
+
x2 + 3
2х
f(x) =
-is f'(x) =
%3D
x2 + 1
[2x² – 1
x2 + 1]
x2 +1
and
(x² + 8)7
(2 – 3x²)5]
Using the product, qoutient and chain rules you learnt in Unit 7 of this Course, could you
(x² + 8)7
14х
30x
that the first derivative of g(x) =
(2 – 3x²)5 is g'(x) =
x² + 8 ' 2 – 3x²]
-
confirm whether his brother is right, wrong or a little tipsy.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13db4cc5-3cbd-4a4f-8c4d-fc00186585e7%2Ffc3f77e3-ff0e-4e5e-a3f5-c1e70c45d5d2%2Fo1wob8i_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, but
currently on some cough medication, believes that the first derivative of
(2x² – 1)(x² + 3)
4x
[(2x² – 1)(x² + 3)
2x
+
x2 + 3
2х
f(x) =
-is f'(x) =
%3D
x2 + 1
[2x² – 1
x2 + 1]
x2 +1
and
(x² + 8)7
(2 – 3x²)5]
Using the product, qoutient and chain rules you learnt in Unit 7 of this Course, could you
(x² + 8)7
14х
30x
that the first derivative of g(x) =
(2 – 3x²)5 is g'(x) =
x² + 8 ' 2 – 3x²]
-
confirm whether his brother is right, wrong or a little tipsy.
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