UNIVERSITY OF SOUTH AFRICA (UNISA)
[College of Science]
⋄
CALCULUS ASSIGNMENT
Assignment 03 — 2026
⋄
Module Code: MAT1512
Module Name: Calculus A
Assignment No.: Assignment 03
Due Date: August 2026
Submitted in partial fulfilment of the requirements for [Module Name]
at the University of South Africa.
, UNISA | MAT1512 Calculus Assignment
Question 1: Differentiate the following:
1(a) Differentiate y = x3 ln x
Answer:
Given:
y = x3 ln x
Use the product rule:
dy dv du
=u +v
dx dx dx
Let
u = x3 , v = ln x
Differentiate each part:
du
= 3x2
dx
dv 1
=
dx x
Substitute into the product rule:
dy 1
= x3 + (ln x)(3x2 )
dx x
Simplify:
= x2 + 3x2 ln x
Factor out x2 :
dy
= x2 (1 + 3 ln x)
dx
1(b) Differentiate y = e2x sin x
Answer:
Given:
y = e2x sin x
Use the product rule.
Page 1 of 12
[College of Science]
⋄
CALCULUS ASSIGNMENT
Assignment 03 — 2026
⋄
Module Code: MAT1512
Module Name: Calculus A
Assignment No.: Assignment 03
Due Date: August 2026
Submitted in partial fulfilment of the requirements for [Module Name]
at the University of South Africa.
, UNISA | MAT1512 Calculus Assignment
Question 1: Differentiate the following:
1(a) Differentiate y = x3 ln x
Answer:
Given:
y = x3 ln x
Use the product rule:
dy dv du
=u +v
dx dx dx
Let
u = x3 , v = ln x
Differentiate each part:
du
= 3x2
dx
dv 1
=
dx x
Substitute into the product rule:
dy 1
= x3 + (ln x)(3x2 )
dx x
Simplify:
= x2 + 3x2 ln x
Factor out x2 :
dy
= x2 (1 + 3 ln x)
dx
1(b) Differentiate y = e2x sin x
Answer:
Given:
y = e2x sin x
Use the product rule.
Page 1 of 12
