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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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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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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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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
=
−
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
=
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
)
=
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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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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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
=
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
=
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
=
e
7
x
5
y=e^{7 x^{5}}
y
=
e
7
x
5
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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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
′
=
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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
′
=
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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
−
5
x
3
y=e^{-5 x^{3}}
y
=
e
−
5
x
3
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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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
′
=
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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
′
=
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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
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
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
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
=
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
=
log
6
(
−
x
5
)
y=\log _{6}\left(-x^{5}\right)
y
=
lo
g
6
(
−
x
5
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
ln
(
−
8
x
5
)
y=\ln \left(-8 x^{5}\right)
y
=
ln
(
−
8
x
5
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
ln
(
4
x
4
)
y=\ln \left(4 x^{4}\right)
y
=
ln
(
4
x
4
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
ln
(
x
6
−
6
x
5
)
y=\ln \left(x^{6}-6 x^{5}\right)
y
=
ln
(
x
6
−
6
x
5
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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Find the derivative of the following function.
\newline
y
=
ln
(
−
2
x
5
)
y=\ln \left(-2 x^{5}\right)
y
=
ln
(
−
2
x
5
)
\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
′
=
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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
′
=
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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
′
=
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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
′
=
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Find the derivative of the following function.
\newline
y
=
ln
(
−
5
x
5
)
y=\ln \left(-5 x^{5}\right)
y
=
ln
(
−
5
x
5
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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- Let
g
g
g
be a function such that
g
(
4
)
=
16
g(4)=16
g
(
4
)
=
16
and
g
′
(
4
)
=
12
g^{\prime}(4)=12
g
′
(
4
)
=
12
.
\newline
- Let
h
h
h
be the function
h
(
x
)
=
x
h(x)=\sqrt{x}
h
(
x
)
=
x
.
\newline
Let
H
H
H
be a function defined as
H
(
x
)
=
g
(
x
)
h
(
x
)
H(x)=\frac{g(x)}{h(x)}
H
(
x
)
=
h
(
x
)
g
(
x
)
.
\newline
H
′
(
4
)
=
H^{\prime}(4)=
H
′
(
4
)
=
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