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Math Problems
Calculus
Find derivatives using logarithmic differentiation
Find the coordinates of the point on the curve
\newline
y
=
2
(
x
−
5
)
x
+
1
y=\frac{2(x-5)}{\sqrt{x+1}}
y
=
x
+
1
2
(
x
−
5
)
where the gradient is
\newline
5
4
\frac{5}{4}
4
5
.
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Question
\newline
The volume of the prism below is given as
120
c
m
3
120 \mathrm{~cm}^{3}
120
cm
3
.
\newline
Use this information to find the length of the side labelled '
z
z
z
'.
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Find the derivative of
y
=
tan
8
x
+
3
(
x
)
y=\tan ^{8 x+3}(x)
y
=
tan
8
x
+
3
(
x
)
.
\newline
Provide vour answer below:
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8
8
8
. Write an equation for the line in slopeintercept form.
\newline
y
=
y=
y
=
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Express the area of the entire rectangle.
\newline
Your answer should be a polynomial in standard form.
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Given
f
(
x
)
=
2
x
−
3
f(x)=2 x-3
f
(
x
)
=
2
x
−
3
, find
f
′
(
−
2
)
f^{\prime}(-2)
f
′
(
−
2
)
using the definition of a derivative.
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3
3
3
The equation of the curve is
y
=
e
2
x
3
+
4
x
y=\frac{e^{2 x}}{3+4 x}
y
=
3
+
4
x
e
2
x
.
\newline
(i) Find the coordinates of the stationary point on the curve, leaving your answer in exact value.
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Find the work done when a force
F
⃗
=
(
x
2
−
y
2
+
2
x
)
i
^
−
(
2
x
y
+
y
)
j
^
\vec{F} = (x^2 - y^2 + 2x)\hat{i}- (2xy + y)\hat{j}
F
=
(
x
2
−
y
2
+
2
x
)
i
^
−
(
2
x
y
+
y
)
j
^
moves a particle in the xy plane from
(
0
,
0
)
(0,0)
(
0
,
0
)
to
(
1
,
1
)
(1,1)
(
1
,
1
)
along the parabola
y
2
=
x
y^2 = x
y
2
=
x
. Is the work done different when the path is the straight line
y
=
x
y = x
y
=
x
?
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Find the area
(
A
)
(A)
(
A
)
and perimeter
(
P
)
(P)
(
P
)
of the enclosed figures.
\newline
A
=
f
t
2
A=\mathrm{ft}^{2}
A
=
ft
2
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Find the volume of the composite solid shown.
\newline
Volume
=
=
=
\newline
Enter your next step here
\newline
c
m
3
\mathrm{cm}^{3}
cm
3
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Find the derivative of
y
=
csc
−
7
x
−
2
(
x
)
y=\csc ^{-7 x-2}(x)
y
=
csc
−
7
x
−
2
(
x
)
. Be sure to include parentheses around the arguments of any logarithmic or trigonometric functions in your answer.
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Region
R
R
R
is enclosed by the line
y
=
x
y=x
y
=
x
and the curve
y
=
x
2
y=x^{2}
y
=
x
2
.
\newline
What is the volume of the solid generated when
R
R
R
is rotated about the
x
x
x
axis?
\newline
Give an exact answer in terms of
π
\pi
π
.
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x
+
y
=
a
,
x
+
z
=
b
,
y
+
z
=
c
x+y=a, x+z=b, y+z=c
x
+
y
=
a
,
x
+
z
=
b
,
y
+
z
=
c
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The curves
x
=
−
1
y
2
−
5
x=-\frac{1}{y^{2}-5}
x
=
−
y
2
−
5
1
and
x
=
1
x=1
x
=
1
are graphed.
\newline
Which expression represents the area bounded by the curves?
\newline
Choose
1
1
1
answer:
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graph of
y
=
5
(
1.2
)
x
y=5(1.2)^x
y
=
5
(
1.2
)
x
is shown in the
−
-
−
plane. Which of the following characteristics of the graph is displayed as a constant or coefficient in the equation as written?
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1
1
1
. (
8
8
8
pts) Using implicit differentiation, find
d
y
/
d
x
d y / d x
d
y
/
d
x
for the curve
\newline
2
cos
(
4
x
3
)
sin
(
7
y
)
=
−
5
x
2 \cos \left(4 x^{3}\right) \sin (7 y)=-5 x
2
cos
(
4
x
3
)
sin
(
7
y
)
=
−
5
x
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If
f
(
x
,
y
)
=
x
y
f(x, y)=x y
f
(
x
,
y
)
=
x
y
, find the gradient vector \(\newlineabla f(6,5) \).
\newline
a
b
l
a
f
(
6
,
5
)
=
⟨
5
,
6
⟩
abla f(6,5)=\langle 5,6\rangle
ab
l
a
f
(
6
,
5
)
=
⟨
5
,
6
⟩
\newline
Use the gradient vector to find the tangent line to the level curve
f
(
x
,
y
)
=
30
f(x, y)=30
f
(
x
,
y
)
=
30
at the point
(
6
,
5
)
(6,5)
(
6
,
5
)
.
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Find the argument of the complex number
−
1
−
3
i
-1-\sqrt{3} i
−
1
−
3
i
in the interval
0
≤
θ
<
2
π
0 \leq \theta<2 \pi
0
≤
θ
<
2
π
. Express your answer in terms of
π
\pi
π
.
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Write an integral expression that will give the length of the path given by
f
(
x
)
=
10
x
4
−
1
f(x) = 10x^4 - 1
f
(
x
)
=
10
x
4
−
1
from
x
=
7
x = 7
x
=
7
to
x
=
8
x = 8
x
=
8
.
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The graph of
y
=
−
3
(
2
)
x
y = -3(2)^x
y
=
−
3
(
2
)
x
is shown in the xy-plane. Which of the following characteristics of the graph is displayed as a constant or coefficient in the equation?
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Follow the steps to compute the volume of the solid obtained by rotating the region bounded by
\newline
x
=
0
,
y
=
1
,
and
x
=
y
7
x=0, \quad y=1, \quad \text { and } \quad x=y^{7}
x
=
0
,
y
=
1
,
and
x
=
y
7
\newline
about the line
y
=
1
y=1
y
=
1
using the method of disks or washers.
\newline
a. Using the method of disks or washers, set up the integral.
\newline
V
=
∫
a
b
V=\int_{a}^{b}
V
=
∫
a
b
\newline
with
a
=
a=
a
=
\newline
and
b
=
b=
b
=
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Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an
(
x
,
y
)
(x, y)
(
x
,
y
)
point.
\newline
y
=
x
2
+
4
y=x^{2}+4
y
=
x
2
+
4
\newline
Answer:
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The derivative of the function
f
f
f
is defined by
f
′
(
x
)
=
x
2
+
5
cos
(
3
x
)
f^{\prime}(x)=x^{2}+5 \cos (3 x)
f
′
(
x
)
=
x
2
+
5
cos
(
3
x
)
. Find the
x
x
x
values, if any, in the interval
−
2
<
x
<
2.5
-2<x<2.5
−
2
<
x
<
2.5
where the function
f
f
f
has a relative maximum. You may use a calculator and round all values to
3
3
3
decimal places.
\newline
Answer:
x
=
x=
x
=
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The derivative of the function
f
f
f
is defined by
f
′
(
x
)
=
x
3
−
5
+
2
sin
(
2
x
−
5
)
f^{\prime}(x)=x^{3}-5+2 \sin (2 x-5)
f
′
(
x
)
=
x
3
−
5
+
2
sin
(
2
x
−
5
)
. Find the
x
x
x
values, if any, in the interval
−
0.5
<
x
<
2.5
-0.5<x<2.5
−
0.5
<
x
<
2.5
where the function
f
f
f
has a relative maximum. You may use a calculator and round all values to
3
3
3
decimal places.
\newline
Answer:
x
=
x=
x
=
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For the following equation, what is the instantaneous rate of change at
x
=
−
2
?
x=-2 ?
x
=
−
2
?
\newline
f
(
x
)
=
2
x
2
−
5
f(x)=2 x^{2}-5
f
(
x
)
=
2
x
2
−
5
\newline
Answer:
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For the following equation, what is the instantaneous rate of change at
x
=
3
x=3
x
=
3
?
\newline
f
(
x
)
=
x
3
−
2
x
−
4
f(x)=x^{3}-2 x-4
f
(
x
)
=
x
3
−
2
x
−
4
\newline
Answer:
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For the following equation, what is the instantaneous rate of change at
x
=
−
2
?
x=-2 ?
x
=
−
2
?
\newline
f
(
x
)
=
−
x
3
−
2
x
f(x)=-x^{3}-2 x
f
(
x
)
=
−
x
3
−
2
x
\newline
Answer:
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For the following equation, what is the instantaneous rate of change at
x
=
−
1
?
x=-1 ?
x
=
−
1
?
\newline
f
(
x
)
=
−
3
x
2
+
2
f(x)=-3 x^{2}+2
f
(
x
)
=
−
3
x
2
+
2
\newline
Answer:
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For the following equation, what is the instantaneous rate of change at
x
=
−
1
?
x=-1 ?
x
=
−
1
?
\newline
f
(
x
)
=
x
5
+
4
x
2
+
4
f(x)=x^{5}+4 x^{2}+4
f
(
x
)
=
x
5
+
4
x
2
+
4
\newline
Answer:
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of people at a carnival is modeled by the following differential equation:
\newline
d
P
d
t
=
2881
7178
P
(
1
−
P
536
)
\frac{d P}{d t}=\frac{2881}{7178} P\left(1-\frac{P}{536}\right)
d
t
d
P
=
7178
2881
P
(
1
−
536
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of people at the carnival is
148
148
148
and is increasing at a rate of
43
43
43
people per hour. At what value of
P
P
P
does the graph of
P
(
t
)
P(t)
P
(
t
)
have an inflection point?
\newline
Answer:
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Find the derivative of the following function.
\newline
y
=
6
−
9
x
5
y=6^{-9 x^{5}}
y
=
6
−
9
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
3
x
3
y=3^{x^{3}}
y
=
3
x
3
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
4
9
x
4
y=4^{9 x^{4}}
y
=
4
9
x
4
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
e
2
x
6
+
2
x
5
y=e^{2 x^{6}+2 x^{5}}
y
=
e
2
x
6
+
2
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
e
5
x
5
y=e^{5 x^{5}}
y
=
e
5
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
e
8
x
6
+
5
x
5
y=e^{8 x^{6}+5 x^{5}}
y
=
e
8
x
6
+
5
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
e
−
6
x
6
−
9
x
5
y=e^{-6 x^{6}-9 x^{5}}
y
=
e
−
6
x
6
−
9
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
e
x
2
−
9
x
y=e^{x^{2}-9 x}
y
=
e
x
2
−
9
x
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
e
−
8
x
5
y=e^{-8 x^{5}}
y
=
e
−
8
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
e
−
5
x
3
y=e^{-5 x^{3}}
y
=
e
−
5
x
3
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
e
7
x
5
y=e^{7 x^{5}}
y
=
e
7
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
e
−
5
x
5
y=e^{-5 x^{5}}
y
=
e
−
5
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
4
2
x
5
y=4^{2 x^{5}}
y
=
4
2
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
4
7
x
4
y=4^{7 x^{4}}
y
=
4
7
x
4
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
log
6
(
−
x
5
)
y=\log _{6}\left(-x^{5}\right)
y
=
lo
g
6
(
−
x
5
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
log
2
(
2
x
2
+
2
x
)
y=\log _{2}\left(2 x^{2}+2 x\right)
y
=
lo
g
2
(
2
x
2
+
2
x
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
ln
(
x
4
)
y=\ln \left(x^{4}\right)
y
=
ln
(
x
4
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
ln
(
4
x
5
)
y=\ln \left(4 x^{5}\right)
y
=
ln
(
4
x
5
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
ln
(
−
2
x
3
)
y=\ln \left(-2 x^{3}\right)
y
=
ln
(
−
2
x
3
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
Find the derivative of the following function.
\newline
y
=
ln
(
−
x
5
)
y=\ln \left(-x^{5}\right)
y
=
ln
(
−
x
5
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
Get tutor help
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