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Math 231
April 6, 2017
1. Use the definition of the derivative to find the instantaneous rate of change of the function f (x) = 5x at the
point x = 2. Show all your work below.
′ = lim
f (2)
x →2
5 x - 25
x-2
x
5 x - 25
x-2
1.95
38.66
1.99
39.91
2
------
2.01
40.56
2.05
41.90
It appears that the derivative at x = 2 is approximately 40.2
You may need to move in closer.
2. Answer any ten (or more) of the following:
a. Is the graph of f(x) = x3 – 4x2 + 5x + 3 increasing or decreasing when x = 2? Give the (calculus) reason
for your answer.
f ′(x) = 3x 2 − 8x + 5 and f ′(2) = 12 − 16 + 5 = 1 > 0 .
Since the derivative is positive when x = 2, the function is increasing there.
3
b. Write the derivative of f (x) = ( sin 5x − cos 7x ) .
2
f ′(x) = 3 ( sin 5x − cos 7x ) ⋅ ( 5 cos 5x + 7 sin 7x )
Math 231
(
April 6, 2017
)
c. Find the instantaneous rate of change of f (x) = ln x 3
x 2 − 4x + 8 at x = 3.
1
x 2 - 4x + 8 = 3ln x + ln ( x 2 - 4x + 8 )
2
3 1 2x - 4
′ = +
f (x)
x 2 x 2 - 4x + 8
12
1
f ′ ( 3 ) = 1+
= 1+ = 1.2
25
5
(
)
f(x) = ln x 3
2
(
)
7
d. Write the derivative of g(x) = e 4x 3x 2 + 1 . (No need to simplify.)
6
g ′(x) = e 4x ⋅ 7 ( 3x 2 + 1) ⋅ 6x + e 4x ⋅ 8x ( 3x 2 + 1)
2
2
7
e. Find the instantaneous rate of change of y at the point (1, 2), where y is determined by the equation
x 2 − 2x + y3 = 7 .
2x − 2 + 3y 2
dy
=0
dx
dy 2 − 2x 2 − 2
=
=
=0
dx
12
3y 2
f. Find a second point on the line that is tangent to y =
The slope of the tangent is given by y ′ = 4 ⋅
Then, when x = 1, the slope is 4 ⋅
at the point (1, 8).
4x − 5
2 2x 2 − 5x + 7
4 −5
2 2−5+ 7
.
= −1 .
Other points on the tangent line include (2, 7), (5, 4), (0, 9), (!3, 12), etc.
The equation of the tangent line is y = ! (x ! 1) + 8 or y = 9 ! x.
Math 231
April 6, 2017
g. At what points is the tangent to y = 2x 3 − 3x 2 − 36x + 10 horizontal?
A horizontal line has a slope of zero. Therefore,
y′ = 6x 2 − 6x − 36 = 6 ( x 2 − x − 6 ) = 6 ( x − 3 )( x + 2 ) = 0
The tangent is horizontal when x = 3 or x = !2, at the points (3, !71) and (!2, 54).
h. Use the derivative rules for sin x and cos x to show that
cot x =
d
(cot x) = − csc 2 x .
dx
cos x
sin x
2
2
d
sin x ⋅ -sin x - cos x ⋅ cos x -sin 2 x - cos 2 x - ( sin x + cos x )
-1
cot x =
=
=
=
= -csc 2 x
2
2
2
2
dx
sin x
sin x
sin x
sin x
i.
Use the definition to prove that
d [ kf (x)]
dx
j.
d[kf (x)]
= kf ′(x) (where k is a constant).
dx
k ( f (x + h) − f (x) )
( f (x + h) − f (x) )
kf (x + h) − kf (x)
= lim
= k lim
= kf ′(x)
h →0
h →0
h →0
h
h
h
= lim
Compare the slope of f(x) = x +
f ′(x) = 1 −
2
x
2
at the point (1, 3).
f ′(1) = 1 − 2 = −1
;
g ′(x) =
2
to that of g(x) =
x
3
2 x
:
g ′(1) =
3
2
The curves are almost, but not quite, perpendicular at (1, 3).
Math 231
k. Find p′(0) for p(x) =
p′(x) =
p′(0) =
l.
4x − 3
.
7x + 10
( 7x + 10 ) ⋅ 4 − ( 4x − 3 ) ⋅ 7 28x + 40 − 28x + 21
61
=
=
2
2
2
( 7x + 10 )
( 7x + 10 )
( 7x + 10 )
61
100
Write the equation of the tangent line to the graph of y = tan −1 x .
1
1
′ = .
f (1)
2
1+ x
2
1
 1

The tangent line equation is y = ( x - 1) + .785  = x + .285 
2
 2

′ =
The slope of the tangent line is given by f (x)
April 6, 2017