MEMORIAL UNIVERSITY OF NEWFOUNDLAND
DEPARTMENT OF MATHEMATICS AND STATISTICS
MATHEMATICS 1000
Assignment 7
SOLUTIONS
[5]
1. First we differentiate both sides of the equation with respect to x:
d
d
[3(x2 + y 2 )2 ] =
[100xy]
dx
dx
d
dy
6(x2 + y 2 ) · [x2 + y 2 ] = 100y + 100x
dx
dx
dy
dy
6(x2 + y 2 ) 2x + 2y
= 100y + 100x
dx
dx
dy
dy
dy
+ 12y 3
− 100x
= 100y − 12x3 − 12xy 2
dx
dx
dx
dy
[12x2 y + 12y 3 − 100x] = 100y − 12x3 − 12xy 2
dx
100y − 12x3 − 12xy 2
dy
=
dx
12x2 y + 12y 3 − 100x
12x2 y
=
25y − 3x3 − 3xy 2
.
3x2 y + 3y 3 − 25x
When x = 1 and y = 3, the slope of the tangent line is
m=
25(3) − 3(13 ) − 3(1)(32 )
45
9
=
= .
2
3
3(1 )(3) + 3(3 ) − 25(1)
65
13
Thus the equation of the tangent line is given by
y−3=
9
(x − 1)
13
=⇒
y=
9
30
x+ .
13
13
Fall 2020
[5]
2. We have
x7 ex
√
ln(y) = ln
tan(x) x2 + 5
√
= ln(x7 ex ) − ln tan(x) x2 + 5
= ln(x7 ) + ln(ex ) − ln(tan(x)) − ln
√
x2 + 5
1
= 7 ln(x) + x − ln(tan(x)) − ln(x2 + 5)
2
d
d
1
2
[ln(y)] =
7 ln(x) + x − ln(tan(x)) − ln(x + 5)
dx
dx
2
1 dy
7
1
1
1
·
= +1−
· sec2 (x) − · 2
· 2x
y dx
x
tan(x)
2 x +5
7
sec2 (x)
x
+1−
− 2
x
tan(x)
x +5
2
dy
7
sec (x)
x
=y
+1−
− 2
dx
x
tan(x)
x +5
7
sec2 (x)
x
x7 ex
√
+1−
− 2
=
.
tan(x)
x +5
tan(x) x2 + 5 x
=