2/24/16
Name
Exam 2 - Derivatives
Find f 0 (x).
1.
f (x) = e2 =) f 0 (x) = 0
The 2’s should be 3’s below
⇡x2 =) f 0 (x) = 2⇡x
2.
f (x) =
3.
f (x) = x2 tan x =) f 0 (x) = 2x tan x + x2 sec2 x
4.
f (x) = 4e
x
9 ln x =) f 0 (x) =
+ sec x
4e
x
+ sec x tan x
9
x
5.
f (x) = (2x4 3e2x +sin x)7 =) f 0 (x) = 7(2x4 3e2x +sin x)6 (8x3 6e2x +cos x)
6.
5
f (x) = p = 5x
x
1/2
=) f 0 (x) = 5
✓
7.
f (x) = (ln x)2 =) f 0 (x) = 2(ln x)
8.
1
2
◆
x
3/2
5
p
2x x
=
✓ ◆
1
2 ln x
=
x
x
f (x) = sec3 x =) f 0 (x) = 3 sec2 x(sec x tan x) = 3 sec3 x tan x
9.
f (x) = esec
3
x
=) f 0 (x) = 3 sec3 x tan xesec
3
x
10.
f (x) = ln(sec3 x) =) f 0 (x) =
3 sec3 x tan x
= 3 tan x
sec3 x
11.
f (x) = x2 sec
1
x =) f 0 (x) = 2x sec
x+x2
✓
(x3 )+x
✓
1
1
x x2
p
1
◆
= 2x sec
1
x+ p
x
x2
1
12.
f (x) = x sin
1
(x3 ) =) f 0 (x) = sin
1
p
13.
f (x) = 37 ln x =) f 0 (x) = 37 ln x (ln 3)
14.
f (x) =
1
1
x6
◆
(3x2 ) = sin
1
(x3 )+ p
✓ ◆
7
(7 ln 3)37 ln x
=
x
x
ex 1
ex (ex + 1) ex (ex
=) f 0 (x) =
ex + 1
(ex + 1)2
1)
=
2ex
(ex + 1)2
15.
f (x) =
x3
3x2 cos x ( sin x)x3
3x2 cos x + x3 sin x
=) f 0 (x) =
=
cos x
cos2 x
cos2 x
16.
f (x) = y = x3x ! ln y = ln(x3x ) = 3x ln x
di↵erentiate ln y = 3x ln x . . .
y0
= 3 ln x + 3x
y
✓ ◆
1
= 3 ln x + 3
x
Now solve for y 0 . . .
f 0 (x) = y 0 = (3 + 3 ln x)y = (3 + 3 ln x)x3x
3x3
1 x6
17.
f (x) = 7x3 tan x tan
f 0 (x) = 21x2 tan x tan
18.
f (x) =
p
x2
19.
1
1
x =)
x + 7x3 sec2 x tan
1 =) f 0 (x) =
1 2
(x
2
1
1/2
1)
x+
7x3 tan x
1 + x2
(2x) = p
x
x2
p
x2 1 sec 1 x =)
p
1
x sec
sec 1 x + s2 1 · p
= p
1
x x2 1
x2
1
f (x) =
f 0 (x) = p
x
x2
20.
f (x) = y =
ln y = ln
3
ln y = ln(x ) + ln e
✓
x3 e4x 1
p
ln(4x 1) x2
x e
p
ln(4x 1) x2
4x 1
ln y = 3 ln x + 4x
3 4x 1
1
ln(ln(4x
ln(ln(4x
1
1))
1))
1
◆
ln
⇣p
1
x 1
+
x
1
⌘
1
x2
1
ln(x2
2
1)
Now di↵erentiate . . .
y0
3
= +4
y
x
y0
3
= +4
y
x
1
ln(4x
·
1) 4x
4
(4x
4
1) ln(4x
1
1
2x
·
2 x2 1
x
1)
x2
1
0
Now solve for y . . .
✓
◆
3
4
x
y0 =
+4
y
x
(4x 1) ln(4x 1) x2 1
✓
◆
3
4
x
x3 e4x 1
p
y0 =
+4
2
x
(4x 1) ln(4x 1) x
1 ln(4x 1) x2
1