C. HECKMAN
265
SOLUTIONS DERIVATIVE MASTERY 1A
√
(1) x 3 5x − 3
√
1
Solution: x · (5x − 3)−2/3 · 5 + 3 5x − 3
3
(2) x2 sec(4x)
Solution: x2 · sec(4x) tan(4x) · 4 + sec(4x) · 2x
(3) y =
1 + cos x
sin x
Solution: 2 ·
2
1 + cos x
sin x
sin x · (− sin x) − (1 + cos x) · cos x
·
sin2 x
(4) 83x−4
Solution: 83x−4 · ln 8 · 3
(5) x2 (8 − 3x)
7/4
7
Solution: x2 · (8 − 3x)3/4 · (−3) + (8 − 3x)7/4 · 2x
4
1
(6) (x2 + 1)(csc x − 5x + 1)
Solution: (x2 + 1)(− csc x cot x − 5) + (csc x − 5x + 1)(2x)
(7)
4
e−x + 4
Solution:
−4 · (−e−x )
(e−x + 4)2
(8) log7 x7 + 8
Solution:
(x7
1
· 7x6
+ 8) · ln 7
(9) arcsin(4x2 + 8x − 9)
1
Solution: p
· (8x + 8)
2
1 − (4x + 8x − 9)2
(10)
√
1 + cos x
Solution:
1
(1 + cos x)−1/2 · (− sin x)
2
2
(11) 5x7 − 3x3/7 + 8x−1/2
Solution: 5 · 7x6 − 3 ·
(12)
3 −4/7
−1 −3/2
x
x
+8·
7
2
tan(ax)
tan(bx)
(where a and b are constants)
Solution:
tan(bx) sec2 (ax) · a − tan(ax) · sec2 (bx) · b
tan2 (bx)
(13) cot(ex )
Solution: − csc2 (ex ) · ex
(14) ln
1
x3
+ ln e
Solution: Note that ln e is a constant, and that ln
easier. −3 ·
1
x
(15) y = e10x sin(20x)
Solution: e10x · cos(20x) · 20 + sin(20x) · e10x · 10
3
1
x3
= ln x−3 = −3 ln x. This made the problem
(16) 3 cos(5x) + 3 sin(x9 )
Solution: 3 · − sin(5x) · 5 + 3 · cos(x9 ) · 9x8
(17) (4x + 5)
π
Solution: π(4x + 5)π−1 · 4
(18) arctan(4ex )
Solution:
(19) Find
1
· 4ex
1 + (4ex )2
dy
where
dx
arccos (x − 2y) + cos (x + 2y) = 1.
Solution:
d
d
[arccos (x − 2y) + cos (x + 2y)] =
[1]
dx
dx
−1
dy
dy
p
· 1−2·
− sin(x + 2y) · 1 + 2 ·
=0
2
dx
dx
1 − (x − 2y)
dy
−1
2
dy
p
− sin(x + 2y) − 2 sin(x + 2y) ·
=0
+p
·
2
2
dx
1 − (x − 2y)
1 − (x − 2y) dx
2
p
1 − (x −
2y)2
·
dy
dy
1
− 2 sin(x + 2y) ·
=p
+ sin(x + 2y)
dx
dx
1 − (x − 2y)2
1
dy
=
dx
p
p
(20) Find
dy
where
dx
Solution:
1 − (x − 2y)2
2
1 − (x − 2y)2
ex+xy + 4y 2 = x.
d x+xy
d
e
+ 4y 2 =
[x]
dx dx
dy
dy
ex+xy · 1 + 1 · y + x ·
+ 8y ·
=1
dx
dx
dy
dy
ex+xy + ex+xy · y + ex+xy · x ·
+ 8y ·
=1
dx
dx
dy
dy
ex+xy · x ·
+ 8y ·
= 1 − ex+xy − ex+xy · y
dx
dx
dy
1 − ex+xy − ex+xy · y
=
dx
ex+xy · x + 8y
4
+ sin(x + 2y)
− 2 sin(x + 2y)
C. HECKMAN
265
SOLUTIONS DERIVATIVE MASTERY 1B
(1)
(2)
1
ln(3x − 2x3 )
3
Solution:
1
1
· (3 − 6x2 )
·
3 3x − 2x3
2
cos x
1 − sin x
Solution: 2 ·
(3) eax ln(b + y)
cos x
1 − sin x
(1 − sin x)(− sin x) − cos x(− cos x)
·
(1 − sin x)2
(where a and b are constants)
Solution: The intended answer was eax ·
lem). The answer eax ·
1
+ ln(b + x) · eax · a (with an x in the original probb+x
1
dy
·
+ ln(b + y) · eax · a was also accepted.
b + y dx
(4) −3 arctan(5 − x2 )
Solution: −3 ·
(5) (2xπ + 8)
1
· (−2x)
1 + (5 − x2 )2
3
Solution: 3(2xπ + 8)2 · 2πxπ−1
1
(6) x2 (3x + 2)
1/4
1
Solution: x2 · (3x + 2)−3/4 · 3 + (3x + 2)1/4 · 2x
4
(7) log2 2x3 + 9
Solution:
(2x3
1
· 6x2
+ 9) ln 2
(8) x3 tan(2x)
Solution: x3 sec2 (2x) · 2 + tan(2x) · 3x2
(9) arccos(9x + 7)
−1
Solution: p
·9
1 − (9x + 7)2
(10)
2 3−e
x
3
Solution:
2
(3 − e)x2−e
3
2
√
(11) x 5 4x + 7
√
1
Solution: x · (4x + 7)−4/5 · 4 + 5 4x + 7
5
(12) arcsin(7x3 − 5)
1
Solution: p
· 7 · 3x2
1 − (7x3 − 5)2
(13) 9x8/3 + 7x3 − 8x−5/2
Solution: 9 ·
8 5/3
−5 −7/2
x + 7 · 3x2 − 8 ·
x
3
2
(14) sec(sec(x2 ))
Solution: sec(sec x2 ) tan(sec x2 ) · sec(x2 ) tan(x2 ) · 2x
(15)
ex + e−x
e
Solution:
ex − e−x
e
3
2
(16) 5x
−8
2
Solution: 5x
(17)
√
−8
· ln 5 · 2x
sin x
Solution:
1
(sin x)−1/2 · cos x
2
(18) (3x5 + 2x3 − 7x + 9)(2x3 + 8x2 − 5x − 7)
Solution: (3x5 + 2x3 − 7x + 9)(6x2 + 16x − 5) + (2x3 + 8x2 − 5x − 7)(15x4 + 6x2 − 7)
(19) Find
dy
where
dx
2 csc x2 y 3 − 5x4 y 3 = 2.
Solution:
d
d
[2 csc x2 y 3 − 5x4 y 3 ] =
[2]
dx
dx
dy
+ y 3 · 2x
2 · − csc(x2 y 3 ) cot(x2 y 3 ) · x2 · 3y 2 ·
dx
4
2 dy
3
3
− 5x · 3y ·
+ y · 20x = 0
dx
dy
+ 2 · − csc(x2 y 3 ) cot(x2 y 3 ) · y 3 · 2x
−2 csc(x2 y 3 ) cot(x2 y 3 ) · x2 · 3y 2 ·
dx
dy
−5x4 · 3y 2 ·
− y 3 · 20x3 = 0
dx
dy
dy
−2 csc(x2 y 3 ) cot(x2 y 3 ) · x2 · 3y 2 ·
− 5x4 · 3y 2 ·
= 2 · csc(x2 y 3 ) cot(x2 y 3 ) · y 3 · 2x + y 3 · 20x3
dx
dx
dy
2 · csc(x2 y 3 ) cot(x2 y 3 ) · y 3 · 2x + y 3 · 20x3
=
dx
−2 csc(x2 y 3 ) cot(x2 y 3 ) · x2 · 3y 2 − 5x4 · 3y 2
(20) Find
dy
where
dx
y 8 − cot (x + 8y) = 2.
Solution:
d 8
d
[y − cot (x + 8y)] =
[2]
dx
dx
dy
dy
8y 7 ·
− − csc2 (x + 8y) · 1 + 8 ·
=0
dx
dx
dy
dy
8y 7 ·
+ csc2 (x + 8y) + csc2 (x + 8y) · 8 ·
=0
dx
dx
dy
dy
8y 7 ·
+ csc2 (x + 8y) · 8 ·
= − csc2 (x + 8y)
dx
dx
dy
− csc2 (x + 8y)
=
dx
8y 7 + csc2 (x + 8y) · 8
4