(1) Suppose
f(x)
=
![1/(2x)](HTMLFiles/2F01sol_1.gif)
. Use
the
limit definition of the derivative
to find
f '(
x).
f '(x) =
![Underscript[lim , h0]](HTMLFiles/2F01sol_2.gif)
![(f(x + h) - f(x))/h](HTMLFiles/2F01sol_3.gif)
=
![Underscript[lim , h 0]](HTMLFiles/2F01sol_4.gif)
![(1/(2 (x + h)) - 1/(2x))/h](HTMLFiles/2F01sol_5.gif)
=
![Underscript[lim , h0]](HTMLFiles/2F01sol_6.gif)
![(x - (x + h))/(2 (x + h) x) /h](HTMLFiles/2F01sol_7.gif)
=
=
![( - h)/( 2 (x + h) x) 1/h](HTMLFiles/2F01sol_11.gif)
=
![( - 1)/( 2 (x + h) x)](HTMLFiles/2F01sol_13.gif)
=
(2)
Find the derivatives of the following functions. You may use any applicable
rule.
(a) f(
x)
= 4
![x^10](HTMLFiles/2F01sol_15.gif)
+
3
![x^3](HTMLFiles/2F01sol_16.gif)
+
![π^2](HTMLFiles/2F01sol_17.gif)
f '(x) =
40
+
9
(b) f(x)
=
f ' (x) =
(c) y
=
x tan(
x)
dy/dx = (1)tan x + x
![sec^2x](HTMLFiles/2F01sol_22.gif)
=
tan x + x ![sec^2x](HTMLFiles/2F01sol_23.gif)
(d) ![d/dx](HTMLFiles/2F01sol_24.gif)
[
25 + cos(
![x ^4](HTMLFiles/2F01sol_25.gif)
)
] = 0 +
![d/dx](HTMLFiles/2F01sol_26.gif)
[ cos(
![x ^4](HTMLFiles/2F01sol_27.gif)
)
] = -sin (
![x ^4](HTMLFiles/2F01sol_28.gif)
)
4
![x^3](HTMLFiles/2F01sol_29.gif)
=
-4
sin
(
)
(e)
[
] =
[
] = 1/2
![(x^2 + 4)^(-1/2)](HTMLFiles/2F01sol_36.gif)
(2x)
=
![x/(x^2 + 4)^(1/2)](HTMLFiles/2F01sol_37.gif)
(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 +
![x^2](HTMLFiles/2F01sol_40.gif)
+
f(
x). Find
h'(2).
h '(x) = 2x + 2x f(x) +
![x^2](HTMLFiles/2F01sol_42.gif)
f
'(x),
so h '(2) = 2(2) + 2(2)(f(2)) +
![2^2](HTMLFiles/2F01sol_43.gif)
f
'(2) = 2(2) + 2(2)(1.5) +
![2^2](HTMLFiles/2F01sol_44.gif)
(-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.
![[Graphics:HTMLFiles/2F01sol_45.gif]](HTMLFiles/2F01sol_45.gif)
(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 +
![1/(2x^(1/2))](HTMLFiles/2F01sol_47.gif)
,
thus the slope we seek is f '(4) = 1 +
![1/(24^(1/2))](HTMLFiles/2F01sol_48.gif)
=
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
![tan(xy) = y^2](HTMLFiles/2F01sol_50.gif)
at the point (π/4, 1).
![dy/dx | _ ( (x, y) = (π/4, 1)) = -sec^2(π/4) /(sec^2(π/4) π/4 - 2) = -2/(2 (π/4) - 2) = -1/((π/4) - 1) =](HTMLFiles/2F01sol_57.gif)
![-4/(π - 4)](HTMLFiles/2F01sol_58.gif)
=
(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
![dx/dt](HTMLFiles/2F01sol_60.gif)
=
2, and we want to find
![dy/dt](HTMLFiles/2F01sol_61.gif)
.
By the Pythagorean Theorem,
![x^2 + y^2 = 10^2](HTMLFiles/2F01sol_62.gif)
.
Differentiating both sides with respect to t:
2x
![2x + ydy/dt = 0](HTMLFiles/2F01sol_65.gif)
, so
Now use the Pythagorean Theorem again to find that x = 8 when y = 6.
Then
![dy/dt = (-2x)/y](HTMLFiles/2F01sol_67.gif)
=