Math 131. Test 2, Form 311, Friday, October 31 Name. Instructions

Math 131.
Test 2, Form 311, Friday, October 31
Name.
Instructions. Do each of the following 8 questions. Question 1 is worth 10 pts, the remaining
questions are worth 5 points each. Please show all appropriate work, and do your best! Express
answers in exact form, unless asked to round. You may use a scientific non-graphing calculator.
1. State the requested derivative rules.
Solution: (a) Constant Multiple rule:
d
[cg(x)] = cg 0 (x)
dx
(where c is a constant)
h0 (x)g(x) − g 0 (x)h(x)
d h(x)
=
(b) Quotient rule:
dx g(x)
g 2 (x)
(c) Difference rule :
(d) Power rule:
d k
[x ] = kxk−1
dx
(e) Product rule:
(f) Chain rule:
d
[g(x) − h(x)] = g 0 (x) − h0 (x)
dx
(where k is a rational number)
d
[g(x)h(x)] = g 0 (x)h(x) + h0 (x)g(x)
dx
d
[h(g(x))] = h0 (g(x)) · g 0 (x)
dx
(g)
d
[cos x] = − sin x
dx
(h)
d
[tan x] = sec2 x
dx
(i)
d
[csc x] = − csc x cot x
dx
(j)
d
[sec x] = sec x tan x
dx
√
2. Let f (x) = 17x + x3
√
6 x
− πx .
x7
(a) Rewrite f (x) using exponents to make its derivative easy to compute:
√
√
17x1/2 , and distribute the x3 and then use properties of exponents
√
√
√
1/2
36 x
3
17x1/2 + 6x−7/2 − πx4
f (x) = 17x + x
−
x
(πx)
=
x7
Solution: (a) First,
to obtain
17x =
(b) Find f 0 (x)
Solution: (b) Now using basic rules for differentiation
√
17 −1/2
0
f (x) =
x
− 21x−9/2 − 4πx3
2
3. (a) State the limit definition for the derivative of a function f (x).
(b) Let f (x) = 7x2 − 6x. Use the definition of derivative to find f 0 (x).
Solution: (a) The derivative of f at x is
f (x + h) − f (x)
h→0
h
f 0 (x) = lim
provided the limit exists.
(b)For f (x) = 7x2 − 6x, we compute
f 0 (x) =
f (x + ∆x) − f (x)
∆x→0
∆x
lim
=
7(x + ∆x)2 − 6(x + ∆x) − (7x2 − 6x)
∆x→0
∆x
=
7(x2 + 2x∆x + (∆x)2 ) − 6x − 6∆x − 7x2 + 6x
∆x→0
∆x
=
(∆x)(14x + 7∆x − 6)
∆x→0
∆x
=
lim
lim
lim
lim 14x + 7∆x − 6 = 14x − 6.
∆x→0
4. Let f (x) = 3 sec x + 5x7 .
(a) Find f 0 (x)
(b) Find f 00 (x)
Solution: (a) f 0 (x) = 3 sec x tan x + 35x6
(b) Then
f 00 x = 3(sec x tan x)(tan x) + 3(sec x)(sec2 x) + 210x5
= 3 sec x tan2 x + 3 sec3 x + 210x5
5. A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is
attached to the front of the boat, which is 5 feet below the level of the pulley. If the rope is pulled
through the pulley at a rate of 24 ft/min, at what rate will the boat be approaching the dock when
60 ft of rope is out? Round your final answer to four decimal places, and state the units of measure
in your answer.
Solution: Let x be the horizontal distance of the front of the boat from the dock and let r be
the length of rope that is out. Then by the pythagorean theorem 52 + x2 = r2 . Differentiating
this equation with respect to t we find
2x
dx
dr
dx
r dr
= 2r
and so
= ·
dt
dt
dt
x dt
dr
From the information given in the question, we know
= −24 ft/min (the negative is because
dt
dx
the rope is being pulled in). We are asked to find
when r = 60 feet. With this information,
dt
from the pythagorean equation above, we know
p
x = 602 − 52 ≈ 59.7913
and therefore,
dx
60
=√
· (−24) ft/min. ≈ −24.0838 ft/min.
dt
602 − 52
The negative sign indicates that the boat is approaching the dock. Thus, the boat is approaching
the dock at a rate of approximately 24.0838 feet per minute.
6. Let h(x) =
5x5 − 6
5x + 8
2
. Find h0 (x), do not simplify your answer.
Solution: First write g(x) =
5x5 − 6
. Then
5x + 8
g 0 (x) =
=
=
25x4 (5x + 8) − (5x5 − 6)(5)
(5x + 8)2
125x5 + 200x4 − 25x5 + 30
(5x + 8)2
5
100x + 200x4 + 30
(5x + 8)2
Therefore,
h0 (x) = 2(g(x))1 g 0 (x)
5
1
100x5 + 200x4 + 30
5x − 6
= 2
5x + 8
(5x + 8)2
2(5x5 − 6)1 (100x5 + 200x4 + 30)
=
(5x + 8)3
√
√
y = 7 and y(16) = 9. Find y 0 (16) by implicit differentiation.
√
√
(b) Find the equation of the tangent line to the graph of x + y = 7 at the point (16, 9). Write
your answer in slope intercept form.
7. (a) Suppose
x+
Solution: (a) Differentiating
√
x+
√
y = 7 implicitly we obtain
1 −1/2 1 −1/2 0
x
+ y
y =0
2
2
or
√
y
dy
x−1/2
= − −1/2 = − √
dx
x
y
√
3
9
Thus y (16) = − √ = − .
4
16
0
(b) Using point slope form with m = −3/4 and the point (16, 9), we write y − 9 = − 43 (x − 16)
and so
3
(3)(16)
3
y =− x+
+9
or
y = − x + 21
4
4
4
8. For what values of x is the tangent line of the graph of f (x) =
x3
− 3x2 − 16x + 1 horizontal?
3
Solution: First, f 0 (x) = x2 − 6x − 16. There is a horizontal tangent when f 0 (x) = 0 and so we
solve x2 − 6x − 16 = 0, thus
x2 − 6x − 16 = (x − 8)(x + 2) = 0.
Thus x = 8 and x = −2 are the x-coordinates where the graph has a horizontal tangent line.

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