(1) Suppose
f(x)
=
. Use
the
limit definition of the derivative
to find
f '(
x).
f '(x) =
=
=
=
=
=
=
(2)
Find the derivatives of the following functions. You may use any applicable
rule.
(a) f(
x)
= 4
+
3
+
f '(x) =
40 +
9
(b) f(x)
=
f ' (x) =
(c) y
=
x tan(
x)
dy/dx = (1)tan x + x
=
tan x + x
(d) [
25 + cos(
)
] = 0 +
[ cos(
)
] = -sin (
)
4
=
-4sin
( )
(e) [
] = [
] = 1/2
(2x)
=
(3) Suppose
f(
x)
equals the number of dollars it costs to erect an
x-foot-high
transmitting tower.
(a) What are the units of
f
'(
x)?
dy/dx =
dollars per foot
(b) Suppose that
f
'(100) = 105. Explain, in ordinary English, what this
means.
It will cost $105 to build the 101st foot
(4) This problem concerns the function
f that is graphed below
(a) Sketch the graph
of
f '(
x).
(Use the same coordinate axis)
(b) Suppose
g(
x)
= sin(
f(
x)).
Find
g '(4).
By the chain rule, g '(x) = cos( f(x) ) f '(x),
so g '(4) = cos(f(4)) f '(4) = cos(1)(0) =
0
(c) Suppose
h(
x)
= 4 +
+
f(
x). Find
h'(2).
h '(x) = 2x + 2x f(x) +
f
'(x),
so h '(2) = 2(2) + 2(2)(f(2)) +
f
'(2) = 2(2) + 2(2)(1.5) +
(-1)
= 4 + 6 - 4 =
6
(5) Sketch the graph of a function
f whose
derivative has the following properties:
f(0) = 2,
f
'(0) = 0,
f
'(3) = 0, and
f
'(
x) ≤ 0 for all values of
x.
(6) Consider the function
f(
x)
=
x +
(a) Find the slope of the tangent line
to the graph of
f at the point where
x = 4.
f '(x) = 1 +
,
thus the slope we seek is f '(4) = 1 +
=
1+1/4 =
5/4
(b) Find the equation of the tangent
line to the graph of
f at the point
where
x = 4.
The slope is 5/4 (from part a), and the line passes through the point (4, f(4))
= (4, 6).
Using the point-slope formula:
y - 6 = 5/4(x - 4)
y - 6 = 5/4 x - 5
y = 5/4 x + 1
(7) Find all values of
x
for which the slope of the tangent to the graph of
y
= sin
x at the point
x is
Slope = dy/dx = cos x, so we are looking for all values of x for which cos x
= 1/2.
Looking at the unit circle, we see that these values of x are x = π/3 + k2π and
x = -π/3 + k2π , where k is an integer.
(8) Find the slope of the
tangent to the graph of
at the point (π/4, 1).
=
(9) Suppose a 10-foot-long ladder is
sliding down a wall in such a way that the base of the ladder moves away from
the wall at a constant rate of 2 feet per second. How fast is the
top of the ladder moving down the wall when it is 6 feet above the floor?
Let x be the distance from the wall to the base of the ladder. Let y be the
height of the top of the ladder.
We know
=
2, and we want to find
.
By the Pythagorean Theorem,
.
Differentiating both sides with respect to t:
2x
, so
Now use the Pythagorean Theorem again to find that x = 8 when y = 6.
Then
=