1. The graph of a function f is given above. Answer the question:
a. Find the value(s) of x where f is not differentiable.
Ans: x = −4, x = −3, x = −2, x = −1, x = 1, x = 2, x = 3, x = 4, x = 5
b. Find the value(s) of x where f is not differentiable but is continuous.
Ans: x = −2, x = 3
c. Find the value(s) of x where f is not continuous, but has a limit.
Ans: x = 1, x = 4
d. Find the value(s) of x where f does not have a limit.
Ans: x = −4, x = −3, x = −1, x = 2, x = 5
2. The graph of a function f is given above. Answer the question:
a. Find the value(s) of x where f is not differentiable.
Ans: x = −5, x = −4, x = −3, x = −2, x = −1, x = 1, x = 2, x = 3 x = 4, x = 5
b. Find the value(s) of x where f is not differentiable but is continuous.
Ans: x = −1, x = 2, x = 3, x = 4
c. Find the value(s) of x where f is not continuous, but has a limit.
Ans: x = −5, x = −3, x = 1
d. Find the value(s) of x where f does not have a limit.
Ans: x = −4, x = −2, x = 5
3. 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
4
9
Ans: f 0 (x) = x2
4
√
e. f (x) = x − 1
d. f (x) =
1
Ans: f 0 (x) = √
2 x
√
√
√
f. f (x) = − x5 + 3 x3 + 7 x
5
9√
7
Ans: f 0 (x) = − x3/2 +
x+ √
2
2
2 x
g. f (x) = ex
Ans: f 0 (x) = ex
h. f (x) = e
Ans: f 0 (x) = 0
i. f (x) = 3xex
Ans: f 0 (x) = 3ex + 3xex
√
x+1
j. y =
3x − 4
√
√
1
(3x − 4)( 2√
) − ( x + 1)(3)
dy
−3x − 6 x − 4
x
√
Ans:
=
=
dx
(3x − 4)2
(3x − 4)2 (2 x)
k. y = (4x + 3)3
Ans:
dy
= 3(4x + 3)2 (4) = 12(4x + 3)2
dx
l. y = (x2 − 3x − 2)5
dy
= 5(x2 − 3x − 2)4 (2x − 3)
dx
p
m. f (x) = 4x3 − 2x2 + 4x + 5
Ans:
12x2 − 4x + 4
6x2 − 2x + 2
Ans: f 0 (x) = √
= √
2 4x3 − 2x2 + 4x + 5
4x3 − 2x2 + 4x + 5
n. f (x) = e3x−1
Ans: f 0 (x) = 3e3x−1
o. y = ex
2
2
dy
= 2xex
dx
p. y = ln x
Ans:
Ans:
dy
1
=
dx
x
q. y = ln(x2 + 4x − 1)
dy
2x + 4
= 2
dx
x + 4x − 1
1
r. y = ln
x
dy
1
Ans:
=−
dx
x
s. y = ln(ln x)
Ans:
dy
1
=
dx
x ln x
1
t. y =
ln x
dy
1
Ans:
=−
dx
x(ln x)2
Ans:
u. y = sin x
Ans: f 0 (x) = cos x
3π
v. f (x) = sin
4
Ans: f 0 (x) = 0
w. f (x) = 3x sin x + sin x cos x
Ans: f 0 (x) = 3 sin x + 3x cos x + cos2 x − sin2 x
x. f (x) = 3x2 ex + 4x tan x
Ans: f 0 (x) = 6xex + 3x2 ex + 4x sec2 x + 4 tan x
y. f (x) = sin x cos xex
Ans: f 0 (x) = sin x cos xex − sin2 xex + cos2 xex
z. y = 3x2 sin x cos xex
dy
= 6x sin x cos xex + 3x2 cos2 xex − 3x2 sin2 xex + 3x2 sin x cos xex
dx
2
aa. y = e4 sin(x −1)
Ans:
2
2
dy
= e4 sin(x −1) 4 cos x2 − 1 2x = 8xe4 sin(x −1) cos x2 − 1
dx
3ex − sin x
bb. y =
cos x − 4ex
Ans:
Ans:
dy
(cos x − 4ex )(3ex − cos x) − (3ex − sin x)(− sin x − 4ex )
(cos x − 4ex )(3ex − cos x) + (3ex − sin x)(sin x + 4ex )
=
=
x
2
dx
(cos x − 4e )
(cos x − 4ex )2
9x2 ex + 4x cos x
√
2 sin x + xex
√ x
√
ex
2 x
dy
(2 sin x + xex )[9(2xex + x2 ex ) + 4(cos x − x sin x)] (9x e + 4x cos x)(2 cos x + 2√x + xe )
√
√
Ans:
=
−
dx
(2 sin x + xex )2
(2 sin x + xex )2
cc. y =
dd. y =
Ans:
ex sin x − x cos x
3x cos x + 4xex
(3x cos x + 4xex )(ex cos x + ex sin x − cos x + x sin x) (ex sin x − x cos x)[3(cos x − x sin x) + 4ex + 4xex ]
dy
=
−
dx
(3x cos x + 4xex )2
(3x cos x + 4xex )2
ee. f (x) = sin(3x2 + 1)
Ans: f 0 (x) = 6x cos(3x2 + 1)
ff. f (x) = cos (sin (4x − 2))
Ans: f 0 (x) = −4 sin(sin(4x − 2))(cos(4x − 2))
gg. y = sin(e3x )
dy
= cos e3x 3e3x
dx
hh. f (x) = sin(4x)
Ans:
Ans: f 0 (x) = 4 cos 4x
ii. y = sin3 x
Ans:
dy
= 3 sin2 x cos x
dx
jj. y = cos4 (x2 + 1)
Ans:
dy
= −8x cos3 (x2 + 1)(sin(x2 + 1))
dx
kk. y = ecos
2
x−1
2
dy
= −2 sin x cos xecos x−1
dx
ll. f (x) = tan3 e2x−1
Ans:
Ans: f 0 (x) = 6 tan2 (e2x−1 )(sec2 (e2x−1 ))(e2x−1 )
mm. f (x) = cos(2x3 − 4)ecos x
Ans: f 0 (x) = ecos x (− sin(2x3 − 4)(6x2 )) + cos(2x3 − 4)(ecos x )(− sin x)
nn. 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
oo. 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
pp. y =
e − 4 sin x cos2 x
(e − 4 sin x cos2 x)[3xex
dy
Ans:
=
dx
qq. y = tan5 sec3 x2 + 3
2
−2
2
(2x) + 3ex −2 ] − (e + 3xex
(e − 4 sin x cos2 x)2
2
−2
)[−4 cos3 x + 8 sin2 x cos x]
dy
= 5 tan4 sec3 x2 + 3 sec2 sec3 x2 + 3 3 sec2 x2 + 3 sec x2 + 3 tan x2 + 3 (2x)
dx
rr. y = arccos x
Ans:
Ans:
1
dy
= −√
dx
1 − x2
ss. y = sin5 e3π+5 + 4e3 x2
Ans:
dy
= 2 sin5 e3π+5 + 4e3 x
dx
tt. y = arcsin(x2 − 5x + 1)
Ans:
dy
2x − 5
=p
dx
1 − (x2 − 5x + 1)2
uu. y = arctan2 (ln 3x)
Ans:
1
dy
= 2 arctan(ln(3x))
dx
x[1 + (ln(3x))2 ]
vv. y =
Ans:
ln(3x)
sin(ln x2 )
sin(ln x2 ) x1 − ln(3x)(cos(ln(x2 )) x2 )
dy
=
dx
(sin(ln(x2 )))2
ww. y =
3e4x sin−1 (3x2 + 1)
ln x sin 5x + 3x2 cos2 x
Ans:
dy
=
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))
1
x
−
sin(5x) + 5 ln x cos(5x) + 3(2x cos2 x − 2x2 cos x sin x)
(ln x sin(5x) + 3x2 cos2 x)2
4. 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 = 2
b. f (x) = 5x − 1, a = −2
Ans: y = 5x − 1
c. f (x) = x2 − 2x + 4, a = 1
Ans: y = 3
d. f (x) = ln x, a = 3
Ans: y − ln 3 =
1
(x − 3)
3
e. f (x) = e3x , a = −2
3
1
= 6 (x + 2)
e6
e
f. f (x) = xex − ln 4x, a = 1
Ans: y −
Ans: y − (e − ln 4) = (2e − 1)(x − 1)
π
g. f (x) = sin x, a =
6
√ 1
3
π
Ans: y − =
x−
2
2
6
5.
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
if
1 ≤ x < 3, x 6= 2
if
x=2
if
3≤x≤5
if
5<x
x−2







5





√

3


x−4





 p
3
(x − 6)4
y
lim f (x) = −3
13
lim f (x) = −2
12
lim f (x) Does not exist.
11
x→−2−
x→−2+
x→−2
lim f (x) = 1, f (0) is undefined.
x→0
10
lim f (x) = 4, f (2) = 5, f is discontinuous at 5.
9
f is continuous at x = 4 but not differentiable at 4.
x→2
8 not second differentiable at 6.
f is first differentiable at x = 6 but
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
6. 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 of a function, f 0 (x), how can you draw the graph of the original function,
b. Given the graph of the derivative
f (x)?
-8
dy
-9
by implicite differentiation:
dx
a. y sin x = x cos y
7. Find
Ans:
dy
cos y − y cos x
=
dx
sin x + x sin y
b. x2 y + y 2 x = 5
Ans:
dy
−2xy − y 2
= 2
dx
x + 2xy
c. 2y − 4x sin x = ey
Ans:
dy
4 sin x + 4x cos x
=
dx
2 − ey
d. ln(xy) +
Ans:
x
=1
y
dy
−y 2 − xy
=
dx
xy − x2
e. exy + ln(x + y) = 2xy
Ans:
dy
2y(x + y) − yexy (x + y) − 1
=
dx
−2x(x + y) + xexy (x + y) + 1
8. 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) = ln x
Ans: f 0 (x) =
e. f (x) =
x2 + 1
x−1
Ans: f 0 (x) =
f. f (x) =
1 00
1
2
f (x) = − 2 f 000 (x) = 3
x
x
x
√
4
12
x2 − 2x − 1 00
f (x) =
f 000 (x) = −
(x − 1)2
(x − 1)3
(x − 1)4
x
1
1
3
Ans: f 0 (x) = √ f 00 (x) = − √ f 000 (x) = √
3
2 x
4 x
8 x5
g. f (x) = x5/4
5 −3/4 000
15
5 1/4 00
x
f (x) =
x
f (x) = − x−7/4
4
16
64
h. f (x) = arccos x
Ans: f 0 (x) =
Ans: f 0 (x) = − √
x
1
f 00 (x) = −
f 000 (x) = −(1 − x2 )−3/2 − 3x2 (1 − x2 )−5/2
2
(1 − x2 )3/2
1−x
i. f (x) = sin x
Ans: f 0 (x) = cos x f 00 (x) = − sin x f 000 (x) = − cos x
j. 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)
9. 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
4
√
√ π
2
2
x−
Ans: L(x) =
−
2
2
4
x
g. f (x) = e , a = 0
f. f (x) = cos x, a =
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