1. Find f (x), or dy dx , as appropriate: a. f(x)=3 Ans: f (x)=0 b. f(x)=4x

1. Find f (x), or dy dx , as appropriate: a. f(x)=3 Ans: f (x)=0 b. f(x)=4x

1. Find f 0 (x), or
dy
, as appropriate:
dx
a. f (x) = 3
Ans: f 0 (x) = 0
b. f (x) = 4x + 1
Ans: f 0 (x) = 4
c. f (x) = −2x2 − 3x + 1
Ans: f 0 (x) = −4x − 3
3x3
d. f (x) =
4
9
Ans: f 0 (x) = x2
4
√
e. f (x) = x − 1
1
Ans: f 0 (x) = √
2 x
√
√
√
5
f. f (x) = − x + 3 x3 + 7 x
7
5
9√
x+ √
Ans: f 0 (x) = − x3/2 +
2
2
2 x
g. y = sin x
Ans: f 0 (x) = cos x
3π
h. f (x) = sin
4
0
Ans: f (x) = 0
i. f (x) = ex
Ans: f 0 (x) = ex
j. f (x) = e
Ans: f 0 (x) = 0
k. f (x) = 3xex
Ans: f 0 (x) = 3ex + 3xex
l. f (x) = 3x sin x + sin x cos x
Ans: f 0 (x) = 3 sin x + 3x cos x + cos2 x − sin2 x
m. f (x) = 3x2 ex + 4x tan x
Ans: f 0 (x) = 6xex + 3x2 ex + 4x sec2 x + 4 tan x
n. f (x) = sin x cos xex
Ans: f 0 (x) = sin x cos xex − sin2 xex + cos2 xex
o. y = 3x2 sin x cos xex
dy
= 6x sin x cos xex + 3x2 cos2 xex − 3x2 sin2 xex + 3x2 sin x cos xex
Ans:
dx
3ex − sin x
p. y =
cos x − 4ex
(cos x − 4ex )(3ex − cos x) − (3ex − sin x)(− sin x − 4ex )
dy
=
Ans:
dx
(cos x − 4ex )2
=
√
(cos x − 4ex )(3ex − cos x) + (3ex − sin x)(sin x + 4ex )
(cos x − 4ex )2
x+1
3x − 4
√
√
(3x − 4)( 2√1 x ) − ( x + 1)(3)
−3x − 6 x − 4
dy
√
=
=
Ans:
dx
(3x − 4)2
(3x − 4)2 (2 x)
3ex − sin x
r. y =
cos x − 4ex
(cos x − 4ex )(3ex − cos x) − (3ex − sin x)(− sin x − 4ex )
dy
=
=
Ans:
dx
(cos x − 4ex )2
q. y =
(cos x − 4ex )(3ex − cos x) + (3ex − sin x)(sin x + 4ex )
(cos x − 4ex )2
9x2 ex + 4x cos x
√
s. y =
2 sin x + xex
√
dy
(2 sin x + xex )[9(2xex + x2 ex ) + 4(cos x − x sin x)]
√
Ans:
=
dx
(2 sin x + xex )2
x
(9x2 ex + 4x cos x)(2 cos x + 2e√x +
√
−
(2 sin x + xex )2
√
xex )
ex sin x − x cos x
t. y =
3x cos x + 4xex
dy
(3x cos x + 4xex )(ex cos x + ex sin x − cos x + x sin x)
Ans:
=
dx
(3x cos x + 4xex )2
(ex sin x − x cos x)[3(cos x − x sin x) + 4ex + 4xex ]
−
(3x cos x + 4xex )2
u. y = (4x + 3)3
dy
Ans:
= 3(4x + 3)2 (4) = 12(4x + 3)2
dx
v. y = (x2 − 3x − 2)5
dy
= 5(x2 − 3x − 2)4 (2x − 3)
dx
p
w. f (x) = 4x3 − 2x2 + 4x + 5
Ans:
6x2 − 2x + 2
12x2 − 4x + 4
= √
Ans: f (x) = √
2 4x3 − 2x2 + 4x + 5
4x3 − 2x2 + 4x + 5
x. f (x) = sin(4x)
0
Ans: f 0 (x) = 4 cos 4x
y. f (x) = e3x−1
Ans: f 0 (x) = 3e3x−1
z. f (x) = sin(3x2 + 1)
Ans: f 0 (x) = 6x cos(3x2 + 1)
aa. f (x) = cos (sin (4x − 2))
Ans: f 0 (x) = −4 sin(sin(4x − 2))(cos(4x − 2))
bb. y = sin(e3x )
dy
Ans:
= cos e3x 3e3x
dx
2
cc. y = ex
dy
2
Ans:
= 2xex
dx
dd. y = sin3 x
dy
Ans:
= 3 sin2 x cos x
dx
ee. y = cos4 (x2 + 1)
dy
Ans:
= −8x cos3 (x2 + 1)(sin(x2 + 1))
dx
2
ff. y = ecos x−1
dy
2
Ans:
= −2 sin x cos xecos x−1
dx
gg. f (x) = tan3 e2x−1
Ans: f 0 (x) = 6 tan2 (e2x−1 )(sec2 (e2x−1 ))(e2x−1 )
hh. f (x) = cos(2x3 − 4)ecos x
Ans: f 0 (x) = ecos x (− sin(2x3 − 4)(6x2 )) + cos(2x3 − 4)(ecos x )(− sin x)
ii. f (x) = x tan(x2 + 1)e3x
Ans: f 0 (x) = tan(x2 + 1)e3x + x[sec2 (x2 + 1)(2x)e3x + 3e3x tan(x2 + 1)]
2
jj. f (x) = x2 esin x + ex sin(cos(3x − 1))
2
2
Ans: f 0 (x) = 2xesin(x ) + x2 (esin(x ) )(cos(x2 ))(2x)+
ex (cos(cos(3x−1)))(− sin(3x−1)(3))+sin(cos(3x−
1))ex
2
e + 3xex −2
kk. y =
e − 4 sin x cos2 x
2
dy
(e − 4 sin x cos2 x)[3xex −2 (2x) + 3ex
Ans:
=
dx
(e − 4 sin x cos2 x)2
2
−2
]
−
(e + 3xex
2
−2
)[−4 cos3 x + 8 sin2 x cos x]
(e − 4 sin x cos2 x)2
2
2x cos(x2 − 4x) + e3x −1
ll. y = 2
3x − 4e3x cos2 (4x+1) + sin3 (5x + 1)
[3x2 − 4e3x cos
dy
=
Ans:
dx
[2x cos(x2 − 4x) + e3x
2
−1
2
(4x+1)
+ sin3 (5x + 1)][2 cos(x2 − 4x) + 2x(− sin(x2 − 4x))(2x − 4) + e3x
(3x2 − 4e3x cos2 (4x+1) + sin3 (5x + 1))2
][6x − 4e3x cos
mm. y = arccos x
dy
1
Ans:
= −√
dx
1 − x2
oo. y = ln x
dy
1
Ans:
=
dx x
pp. y = ln(x2 + 4x − 1)
dy
2x + 4
Ans:
= 2
dx x + 4x − 1
qq. y = arcsin(x2 − 5x + 1)
2x − 5
dy
Ans:
=p
dx
1 − (x2 − 5x + 1)2
rr. y = ln
Ans:
ss. y
Ans:
tt. y
Ans:
1
x
dy
1
=−
dx
x
= ln(ln x)
dy
1
=
dx x ln x
1
=
ln x
dy
1
=−
dx
x(ln x)2
2
(4x+1)
(3 cos2 (4x + 1) + 3x(2 cos(4x + 1))(− sin(4x + 1))(4)
(3x2 − 4e3x cos2 (4x+1) + sin3 (5x + 1))2
uu. y = arctan2 (ln 3x)
1
dy
= 2 arctan(ln(3x))
Ans:
dx
x[1 + (ln(3x))2 ]
vv. y =
ln(3x)
sin(ln x2 )
sin(ln x2 ) x1 − ln(3x)(cos(ln(x2 )) x2 )
dy
=
Ans:
dx
(sin(ln(x2 )))2
3e4x sin−1 (3x2 + 1)
ww. y =
ln x sin 5x + 3x2 cos2 x
dy
Ans:
=
dx
(ln x sin(5x) + 3x2 cos2 x)[12e4x (sin−1 (3x2 + 1) + 3e4x
√
6x
1−(3x2 +1)2
(ln x sin(5x) + 3x2 cos2 x)2
(3e4x sin−1 (3x2 + 1)) x1 sin(5x) + 5 ln x cos(5x) + 3(2x cos2 x − 2x2 cos x
−
(ln x sin(5x) + 3x2 cos2 x)2
2. Find the equation of the line tangent to the given function f at the given
point x = a:
a. f (x) = 2, a = 3
Ans: y = 0
b. f (x) = 5x − 1, a = −2
Ans: y = 5x − 1
c. f (x) = x2 − 2x + 4, a = 1
Ans: y = 3
π
d. f (x) = sin x, a =
6
√ 1
3
π
Ans: y − =
x−
2
2
6
e. f (x) = ln x, a = 3
1
Ans: y − ln 3 = (x − 3)
3
3x
f. f (x) = e , a = −2
1
3
Ans: y − 6 = 6 (x + 2)
e
e
x
g. f (x) = xe − ln 4x, a = 1
Ans: y − (e − ln 4) = (2e − 1)(x − 1)
3. a. Define a function that has a left and right handed limit at a point a, but f
does not have a limit at a.
b. Define a function that has a limit at a point b but is undefined at b.
c. Define a function that has a limit at point c, is defined at c, but is discontinuous
at c.
d. Define a function that is continuous at point d but is not differentiable at d.
e. Define a function that is first differentiable at point p but is not second
differentiable at point p.
f. Draw the graph of a function which has all of the properties in a, b, c, d, e,
above.
Ans: There are more than one correct answer. The function defined below satisfies all conditions a through e:
f (x) =























−3
if
x ≤ −2
−2
if
− 2 < x < −1
sin x
x
if
−1≤x<1
x2 −4
x−2
if
1 ≤ x < 3, x 6= 2
if
x=2
if
3≤x≤5
if
5<x







5





√


3

x−4






p
3
(x − 6)4
lim f (x) = −3
x→−2−
lim f (x) = −2
x→−2+
lim f (x) Does not exist.
x→−2
lim f (x) = 1, f (0) is undefined.
x→0
lim f (x) = 4, f (2) = 5, f is discontinuous at 5.
x→2
f is continuous at x = 4 but not differentiable at 4.
f is first differentiable at x = 6 but not second differentiable at 6.
The graph of f is drawn below.
7
6
5
4
3
2
1
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
-1
-2
-3
-4
-5
4. Think About:
-6
a. Given the graph of the original function f (x), how can you draw the graph of
its derivative, f 0 (x)? -7
0
b. Given the graph of the
-8 derivative of a function, f (x), how can you draw the
graph of the original function, f (x)?
-9
dy
by implicite differentiation:
5. Find
dx
a. y sin x = x cos y
dy
cos y − y cos x
Ans:
=
dx
sin x + x sin y
b. x2 y + y 2 x = 5
dy
−2xy − y 2
Ans:
= 2
dx
x + 2xy
c. 2y − 4x sin x = ey
dy
4 sin x + 4x cos x
Ans:
=
dx
2 − ey
x
d. ln(xy) + = 1
y
dy
−y 2 − xy
Ans:
=
dx
xy − x2
e. exy + ln(x + y) = 2xy
dy
2y(x + y) − yexy (x + y) − 1
Ans:
=
dx −2x(x + y) + xexy (x + y) + 1
6. Find f 00 (x) and f 000 (x):
a. f (x) = 4
Ans: f 0 (x) = 0 f 00 (x) = 0 f 000 (x) = 0
b. f (x) = 3x2
Ans: f 0 (x) = 6x f 00 (x) = 6 f 000 (x) = 0
c. f (x) = ex
Ans: f 0 (x) = ex f 00 (x) = ex f 000 (x) = ex
d. f (x) = sin x
Ans: f 0 (x) = cos x f 00 (x) = − sin x f 000 (x) = − cos x
e. f (x) = ln x
1 00
1
2
f (x) = − 2 f 000 (x) = 3
x
x
x
f. f (x) = arccos x
x
1
f 00 (x) = −
Ans: f 0 (x) = − √
f 000 (x) = −(1 − x2 )−3/2 − 3x2 (1 −
2
3/2
2
(1 − x )
1−x
2 −5/2
x)
Ans: f 0 (x) =
x2 + 1
g. f (x) =
x−1
x2 − 2x − 1 00
4
12
000
Ans: f 0 (x) =
f
(x)
=
f
(x)
=
−
(x − 1)2
(x − 1)3
(x − 1)4
h. f (x) = esin x
Ans: f 0 (x) = cos xesin x f 00 (x) = esin x (cos2 x − sin x) f 000 (x) = esin x (cos3 x −
3 sin x cos x − cos x)
√
i. f (x) = x
1
1
3
Ans: f 0 (x) = √ f 00 (x) = − √ f 000 (x) = √
2 x
4 x3
8 x5
j. f (x) = x5/4
5
5
15
Ans: f 0 (x) = x1/4 f 00 (x) = x−3/4 f 000 (x) = − x−7/4
4
16
64
7. Find the linearization of the given function at the given point a:
a. f (x) = 2x − 3, a = 2
Ans: L(x) = 2x − 3, the linearization of a line is itself.
b. f (x) = 2x − 3, a = 4
Ans: L(x) = 2x − 3, the linearization of a line is itself.
c. f (x) = 2x − 3, a = 10
Ans: L(x) = 2x − 3, the linearization of a line is itself.
d. f (x) = 4x2 − x + 3, a = −1
Ans: L(x) = −9x − 1
e. f (x) = sin x, a = 0
Ans: L(x) = x
π
f. f (x) = cos x, a =
4
√
√ π
2
2
Ans: L(x) =
x−
−
2
2
4
x
g. f (x) = e , a = 0
Ans: L(x) = x + 1
h. f (x) = ex , a = 1
Ans: L(x) = e + e(x − 1)
i. f (x) = ln x, a = 1
Ans: L(x) = x − 1
j. f (x) = ln x, a = 2
1
Ans: L(x) = (ln 2) + (x − 2)
2
k. f (x) = arctan x, a = 0
Ans: L(x) = x