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MAT2691 ASSIGNMENT 1 SEMESTER 1 2024
- Institution
- University Of South Africa (Unisa)
This document contains MAT2691 ASSIGNMENT 1 SEMESTER 1 2024 solutions. Clear step by step calculations are provided.
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MAT2691ASSIGNMENT 1
SEMESTER 1
2024
, QUESTION 1
Solution:
1.1) .
sin x2 tan4 x
y=
x2 + 1 2
Let: u = sin x2 tan4 x and v = x2 + 1 2
By Product rule:
du d d
= sin x2 tan4 x + tan4 x sin x2
dx dx dx
= sin x2 ∙ 4tan3 x sec2 x + tan4 x ∙ 2x cos x2
= 4tan3 x sec2 x sin x2 + 2xtan4 x cos x2
= tan3 x 4sec2 x sin x2 + 2x tan x cos x2
v = x2 + 1 2
dv
= 2 x2 + 1 ∙ 2x
dx
dv
= 4x x2 + 1
dx
By Quotient rule:
, du dv
dy v dx − u dx
=
dx v2
dy x2 + 1 2
tan3 x 4sec2 x sin x2 + 2x tan x cos x2 − sin x2 tan4 x 4x x2 + 1
=
dx x2 + 1 2 2
dy x2 + 1 x2 + 1 tan3 x 4sec2 x sin x2 + 2x tan x cos x2 − 4xsin x2 tan4 x
=
dx x2 + 1 4
dy x2 + 1 tan3 x 4sec2 x sin x2 + 2x tan x cos x2 − 4xsin x2 tan4 x
=
dx x2 + 1 3
1.2) .
2
ex cos ex
y=
sin−1 4x
2
Let: u = ex cos ex and v = sin−1 4x
du 2 d d x2
= ex cos ex + cos ex e
dx dx dx
du 2 2
= ex − ex sin ex + cos ex 2xex
dx
du 2
= ex 2xcos ex − ex sin ex
dx
v= sin−1 4x
1
v = sin−1 4x 2
dv 1 −
1 1
= sin−1 4x 2 ∙ ∙4
dx 2 1 − 4x 2
dv 2
=
dx sin−1 4x 1 − 16x2
dv 2
=
dx 1 − 16x2 sin−1 4x
By Quotient rule:
, du dv
dy v dx − u dx
=
dx v2
2 2 2
sin−1 4x ex 2xcos ex − ex sin ex − ex cos ex
dy 1 − 16x2 sin−1 4x
= 2
dx
sin−1 4x
2 1
−
x2 x x x 2ex cos ex 1 − 16x2 2
e sin−1 4x 2xcos e − e sin e −
dy sin−1 4x
= 2
dx
sin−1 4x
2 2 1
−
ex sin−1 4x 2xcos ex − ex sin ex − 2ex cos ex 1 − 16x2 2
dy sin−1 4x
= 2
dx
sin−1 4x
2 2 1
−
dy ex sin−1 4x 2xcos ex − ex sin ex − 2ex cos ex 1 − 16x2 2
= 3
dx
sin−1 4x
2 1
−
dy ex sin−1 4x 2xcos ex − ex sin ex − 2cos ex 1 − 16x2 2
= 3
dx
sin−1 4x
1.3)
x
y = 4 sec−1 − x2 − 16
4
dy d x d
= 4 sec−1 − x2 − 16
dx dx 4 dx
dy 1 x
=4 4 −
dx x2 − 16
x x 2
−1
4 4
dy 1 x
=4 −
dx x2 − 16
x2
x −1
16
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