MEMORIAL UNIVERSITY OF NEWFOUNDLAND
DEPARTMENT OF MATHEMATICS AND STATISTICS
MATH 1000
Assignment 6
Fall 2015
SOLUTIONS
[4]
1. (a) We use the Product Rule, followed by the Chain Rule:
f 0 (x) = [cos(x)]0 log5 (cos(x)) + cos(x)[log5 (cos(x))]0
= − sin(x) log5 (cos(x)) + cos(x) ·
= − sin(x) log5 (cos(x)) −
[5]
1
[cos(x)]0
cos(x) ln(5)
1
sin(x).
ln(5)
(b) Since this function is of the form [f (x)]g(x) , we must use logarithmic differentiation:
2
ln(y) = ln [sec(x)]x −9
= (x2 − 9) ln(sec(x))
d
d
[ln(y)] =
[(x2 − 9) ln(sec(x))]
dx
dx
1 dy
1
·
= 2x ln(sec(x)) + (x2 − 9) ·
· sec(x) tan(x)
y dx
sec(x)
= 2x ln(sec(x)) + (x2 − 9) tan(x)
dy
= y 2x ln(sec(x)) + (x2 − 9) tan(x)
dx
2
= [sec(x)]x −9 2x ln(sec(x)) + (x2 − 9) tan(x) .
[5]
(c) First we simplify the function using the properties of logarithms:
√
3
x csc5 (x)
f (x) = ln
(3x − 1)7
√
= ln 3 x csc5 (x) − ln (3x − 1)7
√ = ln 3 x + ln csc5 (x) − ln (3x − 1)7
=
1
ln(x) + 5 ln(csc(x)) − 7 ln(3x − 1).
3
Now we differentiate:
1 1
1
1
· +5·
· [csc(x)]0 − 7 ·
· [3x − 1]0
3 x
csc(x)
3x − 1
1
5
7
=
+
· [− csc(x) cot(x)] −
·3
3x csc(x)
3x − 1
1
21
=
− 5 cot(x) −
.
3x
3x − 1
f 0 (x) =
[4]
(d) We use the Chain Rule twice (or, alternatively, the Chain Rule followed by the Quotient
Rule):
f 0 (x) =
1+
=
1+
1
−1 0
2 · [(ln(x)) ]
1
−2
0
2 · −(ln(x)) · [ln(x)]
1
ln(x)
1
ln(x)
−1
1
·
[ln(x)]2 + 1 x
−1
.
=
x ([ln(x)]2 + 1)
=
[4]
(e) We use the Chain Rule twice:
dy
d
= − sin(cosh(arccos(x))) · [cosh(arccos(x))]
dx
dx
d
= − sin(cosh(arccos(x))) sinh(arccos(x)) · [arccos(x)]
dx
−1
= − sin(cosh(arccos(x))) sinh(arccos(x)) · √
1 − x2
sin(cosh(arccos(x))) sinh(arccos(x))
√
=
.
1 − x2
[5]
2. We differentiate both sides of the equation with respect to x, and obtain
d
d 4 3
[x y ] =
[sin(x3 ) − sinh(y 4 )]
dx
dx
dy
dy
4x3 y 3 + 3y 2 x4 = cos(x3 ) · 3x2 − cosh(y 4 ) · 4y 3
dx
dx
dy
dy
3x4 y 2
+ 4y 3 cosh(y 4 )
= 3x2 cos(x3 ) − 4x3 y 3
dx
dx
dy
3x2 cos(x3 ) − 4x3 y 3
= 4 2
.
dx
3x y + 4y 3 cosh(y 4 )
[6]
3. We have
x7 e x
√
ln(y) = ln
tan(x) x2 + 5
√
7 x
2
= ln(x e ) − ln tan(x) x + 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
sec2 (x)
x
7
+1−
− 2
x
tan(x)
x +5
7
dy
sec2 (x)
x
=y
+1−
− 2
dx
x
tan(x)
x +5
sec2 (x)
x
x7 e x
7
√
+1−
− 2
=
.
tan(x)
x +5
tan(x) x2 + 5 x
=
[7]
4. First we have
1
1
y0 = q
· √
√ 2 2 x
1 − ( x)
1
= √ √
2 x 1−x
1
.
= √
2 x − x2
The slope of the tangent line is
√
1
2 3
m=y
=
4
3
0
and so the slope of the normal line is
√
1
3
mN = − = −
.
m
2
Also, note that when x = 41 ,
1
π
y = arcsin
= .
2
6
Thus the equation of the tangent line is
√ π
2 3
1
y− =
x−
6
3
4
√
√
2 3
π− 3
y=
x+
3
6
while the equation of the normal line is
√ 3
1
π
x−
y− =−
6
2
4
√
√
3
3 π
y=−
x+
+ .
2
8
6