Math 2040 Test 2. Fall 2015
1. The graph of y = f (x) is shown above.
where xy 3 + 1 = x4 + y 4 .
(a) For what value(s) x does f (x) fail to be
continuous?
13. Use logarithmic differentiation to find the
derivative of the function f (x) = xtan x .
(b) For what value(s) x is f 0 (x) = 0?
14. Match the function on the top row with
its derivative on the bottom.
(c) For what value(s) x is f 0 (x) undefined?
Questions 2–9: compute the derivative of the
function; simplify your answer.
2.
f (x) =
3.
f (x) =
√
√
x+
1
− 13
x
xex
1
1
1
1
A
1
1
1
1
1
3 5
4.
f (x) = ln((1 + x ) )
5.
f (x) = ln(sec x + tan x)
6.
f (x) = (2x + 1)3 (x − 3)−2
2
B
1
1
1
1
7.
8.
f (x) =
sin x
1 + cos2 x
f (x) = x sin x + cos x
√
9.
C
3
10. Find the equation of the tangent line to
the curve y = 2ex + sin x where x = 0.
11. Evaluate the limit (don’t just write an
answer–show your work):
sin(3x)
.
x→0
4x
lim
12. Use implicit differentiation to find dy/dx
1
1
x
f (x) =
x+1
1
1
4
1
1
1
5
D
1
E
7.
solutions
1. a) x = −4, 2, 3
−4, −2, 2, 3
b) x = 5
c) x =
(1 + cos2 x) cos x − sin x · 2 cos x(− sin x)
(1 + cos2 x)2
cos x + cos3 x + 2 sin2 x cos x
=
(1 + cos2 x)2
cos x(1 + cos2 x + sin2 x + sin2 x)
=
(1 + cos2 x)2
cos x(2 + sin2 x)
=
(1 + cos2 x)2
f 0 (x) =
2.
1
f 0 (x) = x−1/2 − x−2
2
1
1
= √ − 2
2 x x
8.
3.
1
f 0 (x) = x1/2 ex + x−1/2 ex
2
ex
1/2
= (2x + x−1/2 )
2 ex 2x + 1
=
2
x1/2
f 0 (x) = x cos x + sin x − sin x = x cos x
9.
(x + 1)(1/2)x−1/2 − x1/2
(x + 1)2
(x + 1)(1/2) − x
=
(x + 1)2 x1/2
x + 1 − 2x
=
2(x + 1)2 x1/2
1−x
√
=
2(x + 1)2 x
f 0 (x) =
4.
f (x) = 5 ln(1 + x3 )
so
f 0 (x) = 5 ·
1
15x2
2
·
3x
=
.
1 + x3
1 + x3
10.
5.
1
f (x) =
(sec x tan x + sec2 x)
sec x + tan x
sec x(tan x + sec x)
=
sec x + tan x
= sec x
y 0 = 2ex + cos x
0
so the slope is
m = y 0 (0) = 2e0 + cos(0) = 3.
Since
y(0) = 2e0 + sin(0) = 2,
a point on the tangent line is (0, 2). The equation of the tangent line is y = 3x + 2.
6.
f 0 (x) = (2x + 1)3 [−2(x − 3)−3 ]
+ (x − 3)−2 [3(2x + 1)2 · 2]
=
=
=
=
−2(2x + 1)3 6(2x + 1)2
+
(x − 3)3
(x − 3)2
2(2x + 1)2 −(2x + 1)
+3
(x − 3)2
x03
2(2x + 1)2 −2x − 1 + 3x − 9
(x − 3)2
x−3
2
2(2x + 1) (x − 10)
(x − 3)2
11.
sin(3x)
sin(3x) 3
3
= lim
· = .
x→0
x→0
4x
3x
4
4
lim
12.
x · 3y 2 y 0 + y 3 = 4x3 + 4y 3 y 0
3xy 2 y 0 − 4y 3 y 0 = 4x3 − y 3
y=
4x3 − y 3
3xy 2 − 4y 3
13.
y = xtan x
ln y = ln xtan x
ln y = tan x ln x
1 0 tan x
y =
+ ln x · sec2 x
y
x
tan x
2
y=
+ ln x · sec x xtan x
x
14. 1 7→ B, 2 7→ E, 3 7→ A, 4 7→ C, 5 7→ D.