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Math Problems
Calculus
Find derivatives of logarithmic functions
ln
(
x
)
+
ln
(
y
)
−
ln
(
x
y
)
\ln(x) + \ln(y) - \ln(xy)
ln
(
x
)
+
ln
(
y
)
−
ln
(
x
y
)
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Solve for
k
k
k
.
\newline
4
3
=
11
k
k
=
□
\begin{array}{l} \frac{4}{3}=\frac{11}{k} \\ k=\square \end{array}
3
4
=
k
11
k
=
□
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Find the derivative of
y
=
(
9
x
2
−
2
)
sec
(
x
)
y=\left(9 x^{2}-2\right)^{\sec (x)}
y
=
(
9
x
2
−
2
)
s
e
c
(
x
)
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Find the value of
f
(
−
1
)
+
f
(
2
)
−
f
(
4
)
f(-1)+f(2)-f(4)
f
(
−
1
)
+
f
(
2
)
−
f
(
4
)
, where
\newline
f
(
x
)
=
{
2
x
−
4
for
x
≥
4
x
for
0
≤
x
<
4
−
2
for
x
<
0
f(x)=\left\{\begin{array}{lll} \sqrt{2 x-4} & \text { for } x \geq 4 \\ x & \text { for } 0 \leq x<4 \\ -2 & \text { for } x<0 \end{array}\right.
f
(
x
)
=
⎩
⎨
⎧
2
x
−
4
x
−
2
for
x
≥
4
for
0
≤
x
<
4
for
x
<
0
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22
22
22
)
log
3
8
≈
1.9
log
3
10
≈
2.1
log
3
11
≈
2.2
Find
log
3
100
\begin{array}{l}\log _{3} 8 \approx 1.9 \\ \log _{3} 10 \approx 2.1 \\ \log _{3} 11 \approx 2.2 \\ \text { Find } \log _{3} 100\end{array}
lo
g
3
8
≈
1.9
lo
g
3
10
≈
2.1
lo
g
3
11
≈
2.2
Find
lo
g
3
100
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The horsepower,
H
H
H
produced by a truck engine is proportional to the cube of the truck's speed,
S
S
S
.
\newline
Calculate the percentage increase in horsepower that is needed to double the speed.
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lg
x
+
54
log
x
10
=
15
\lg x+54 \log _{x} 10=15
l
g
x
+
54
lo
g
x
10
=
15
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g
(
x
)
=
−
log
5
(
−
x
−
2
)
+
3
g(x)=-\log _{5}(-x-2)+3
g
(
x
)
=
−
lo
g
5
(
−
x
−
2
)
+
3
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Find the derivative of
f
(
x
)
f(x)
f
(
x
)
.
\newline
f
(
x
)
=
x
−
1
f(x)=\sqrt{x-1}
f
(
x
)
=
x
−
1
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You might need: 国 Calculator
\newline
g
(
x
)
=
−
x
2
4
+
7
g(x)=-\frac{x^{2}}{4}+7
g
(
x
)
=
−
4
x
2
+
7
\newline
What is the average rate of change of
g
g
g
over the interval
[
−
2
,
4
]
[-2,4]
[
−
2
,
4
]
?
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Find the derivative of the function
f
(
x
)
=
ln
(
x
2
−
3
x
+
2
)
f(x)=\ln(x^{2}-3x+2)
f
(
x
)
=
ln
(
x
2
−
3
x
+
2
)
.
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b)
lim
x
→
0
sin
(
2
x
)
sin
(
x
)
\lim_{x \rightarrow 0}\frac{\sin(2x)}{\sin(x)}
lim
x
→
0
s
i
n
(
x
)
s
i
n
(
2
x
)
.
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What is the derivative of
(
x
+
1
)
sin
x
(x + 1) \sin x
(
x
+
1
)
sin
x
?
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Determine each sum or difference.
\newline
1
+
(
−
4
)
=
−
3
−
4
+
(
−
1
)
=
−
5
−
3
−
(
−
11
)
=
11
+
(
−
1
)
=
6
+
(
−
10
)
=
5
+
(
−
12
)
=
8
−
(
−
14
)
=
−
8
−
(
−
5
)
=
−
8
+
(
−
5
)
=
\begin{array}{l} 1+(-4)=-3 \\ -4+(-1)=-5 \\ -3-(-11)= \\ 11+(-1)= \\ 6+(-10)= \\ 5+(-12)= \\ 8-(-14)= \\ -8-(-5)= \\ -8+(-5)= \end{array}
1
+
(
−
4
)
=
−
3
−
4
+
(
−
1
)
=
−
5
−
3
−
(
−
11
)
=
11
+
(
−
1
)
=
6
+
(
−
10
)
=
5
+
(
−
12
)
=
8
−
(
−
14
)
=
−
8
−
(
−
5
)
=
−
8
+
(
−
5
)
=
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Solve for the exact value of
x
x
x
.
\newline
log
4
(
9
x
)
−
2
log
4
(
7
)
=
1
\log _{4}(9 x)-2 \log _{4}(7)=1
lo
g
4
(
9
x
)
−
2
lo
g
4
(
7
)
=
1
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
6
(
6
x
)
+
2
log
6
(
4
)
=
1
\log _{6}(6 x)+2 \log _{6}(4)=1
lo
g
6
(
6
x
)
+
2
lo
g
6
(
4
)
=
1
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
2
(
8
x
)
−
3
log
2
(
6
)
=
0
\log _{2}(8 x)-3 \log _{2}(6)=0
lo
g
2
(
8
x
)
−
3
lo
g
2
(
6
)
=
0
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
7
(
8
x
)
+
2
log
7
(
8
)
=
2
\log _{7}(8 x)+2 \log _{7}(8)=2
lo
g
7
(
8
x
)
+
2
lo
g
7
(
8
)
=
2
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
6
(
5
x
)
−
3
log
6
(
5
)
=
1
\log _{6}(5 x)-3 \log _{6}(5)=1
lo
g
6
(
5
x
)
−
3
lo
g
6
(
5
)
=
1
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
4
(
6
x
)
+
2
log
4
(
9
)
=
1
\log _{4}(6 x)+2 \log _{4}(9)=1
lo
g
4
(
6
x
)
+
2
lo
g
4
(
9
)
=
1
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
2
(
2
x
)
+
3
log
2
(
3
)
=
0
\log _{2}(2 x)+3 \log _{2}(3)=0
lo
g
2
(
2
x
)
+
3
lo
g
2
(
3
)
=
0
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
3
(
9
x
)
−
2
log
3
(
3
)
=
2
\log _{3}(9 x)-2 \log _{3}(3)=2
lo
g
3
(
9
x
)
−
2
lo
g
3
(
3
)
=
2
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
2
(
8
x
)
−
2
log
2
(
6
)
=
3
\log _{2}(8 x)-2 \log _{2}(6)=3
lo
g
2
(
8
x
)
−
2
lo
g
2
(
6
)
=
3
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
7
(
6
x
)
+
4
log
7
(
3
)
=
3
\log _{7}(6 x)+4 \log _{7}(3)=3
lo
g
7
(
6
x
)
+
4
lo
g
7
(
3
)
=
3
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
log
6
(
8
x
)
−
3
log
6
(
3
)
=
0
\log _{6}(8 x)-3 \log _{6}(3)=0
lo
g
6
(
8
x
)
−
3
lo
g
6
(
3
)
=
0
\newline
Answer:
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f
(
x
)
=
−
x
2
+
7
x
−
10
6
x
+
11
f(x)=\frac{-x^{2}+7 x-10}{6 x+11}
f
(
x
)
=
6
x
+
11
−
x
2
+
7
x
−
10
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Find derivative
f
(
x
)
=
−
x
2
+
7
x
−
10
6
x
+
11
find derivative
\begin{array}{l}\text { Find derivative } \\ f(x)=\frac{-x^{2}+7 x-10}{6 x+11} \\ \text { find derivative }\end{array}
Find derivative
f
(
x
)
=
6
x
+
11
−
x
2
+
7
x
−
10
find derivative
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The functions
f
f
f
and
g
g
g
are given by
\newline
f
(
x
)
=
1
2
log
10
(
x
−
3
)
g
(
x
)
=
3
ln
(
x
+
2
)
\begin{array}{l} f(x)=\frac{1}{2} \log _{10}(x-3) \\ g(x)=3 \ln (x+2) \end{array}
f
(
x
)
=
2
1
lo
g
10
(
x
−
3
)
g
(
x
)
=
3
ln
(
x
+
2
)
\newline
(A) Solve
f
(
x
)
=
1
f(x)=1
f
(
x
)
=
1
for values of
x
x
x
in the domain of
f
f
f
.
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Solve for
x
x
x
. Leave fractions as fra
\newline
3
log
b
(
2
x
)
+
log
b
(
8
)
=
log
b
(
1
)
x
=
□
\begin{array}{l} 3 \log _{b}(2 x)+\log _{b}(8)=\log _{b}(1) \\ x=\square \end{array}
3
lo
g
b
(
2
x
)
+
lo
g
b
(
8
)
=
lo
g
b
(
1
)
x
=
□
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Find the area under the graph of
f
f
f
over the interval
[
−
2
,
4
]
[-2,4]
[
−
2
,
4
]
.
\newline
f
(
x
)
=
{
x
2
+
6
x
≤
2
5
x
x
>
2
f(x)=\left\{\begin{array}{ll} x^{2}+6 & x \leq 2 \\ 5 x & x>2 \end{array}\right.
f
(
x
)
=
{
x
2
+
6
5
x
x
≤
2
x
>
2
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what is the slope of the line
y
−
4
=
−
7
(
x
−
6
)
y-4=-7(x-6)
y
−
4
=
−
7
(
x
−
6
)
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What value of
y
y
y
is a solution to this equation?
\newline
3
y
=
3
3y = 3
3
y
=
3
\newline
Choices:
\newline
(A)
y
=
1
y = 1
y
=
1
\newline
(B)
y
=
3
y = 3
y
=
3
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Nilai
x
x
x
yang memenuhi persamaan
\newline
sin
2
x
−
4
sin
x
−
5
=
0
\sin ^{2} x-4 \sin x-5=0
sin
2
x
−
4
sin
x
−
5
=
0
\newline
Untuk
0
≤
x
≤
2
π
0 \leq x \leq 2 \pi
0
≤
x
≤
2
π
adalah ....
\newline
(A)
0
0
0
\newline
(B)
π
2
\frac{\pi}{2}
2
π
\newline
(C)
π
\pi
π
\newline
(D)
3
2
π
\frac{3}{2} \pi
2
3
π
\newline
(E)
2
π
2 \pi
2
π
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Find the derivative of
h
(
x
)
=
(
4
x
2
−
8
)
cos
(
x
)
h(x)=\left(4 x^{2}-8\right) \cos (x)
h
(
x
)
=
(
4
x
2
−
8
)
cos
(
x
)
by applying the product rule.
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5
5
5
. Find the derivative of
f
(
x
)
=
16
x
5
4
f(x)=\sqrt[4]{16 x^{5}}
f
(
x
)
=
4
16
x
5
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the derivative of
f
(
x
)
=
16
x
5
4
f(x)=\sqrt[4]{16 x^{5}}
f
(
x
)
=
4
16
x
5
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If
y
(
x
)
=
(
x
x
x
)
,
x
>
0
y(x)=\left(x^{x^{x}}\right), x>0
y
(
x
)
=
(
x
x
x
)
,
x
>
0
then
d
2
x
d
y
2
+
20
\frac{d^{2} x}{d y^{2}}+20
d
y
2
d
2
x
+
20
at
x
=
1
x=1
x
=
1
is equal to:
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3
3
3
.
g
(
x
)
=
ln
(
cos
(
x
3
)
)
g(x)=\ln \left(\cos \left(x^{3}\right)\right)
g
(
x
)
=
ln
(
cos
(
x
3
)
)
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Differentiate
f
(
x
)
=
cos
(
ln
6
x
)
f(x)=\cos(\ln 6x)
f
(
x
)
=
cos
(
ln
6
x
)
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Find
k
′
(
x
)
k'(x)
k
′
(
x
)
.
\newline
k
(
x
)
=
e
x
(
−
x
3
5
)
k(x)=e^{x}(-x^{\frac{3}{5}})
k
(
x
)
=
e
x
(
−
x
5
3
)
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s
(
s
−
1
)
=
2
s(s-1)=2
s
(
s
−
1
)
=
2
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Find all critical points of the function
g
(
θ
)
=
sin
2
(
6
θ
)
.
g(\theta)=\sin^{2}(6\theta).
g
(
θ
)
=
sin
2
(
6
θ
)
.
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Find all critical points of the function
\newline
f
(
x
)
=
cos
−
1
(
x
)
+
3
x
f(x)=\cos^{-1}(x)+3x
f
(
x
)
=
cos
−
1
(
x
)
+
3
x
for
−
1
<
x
<
1.
-1 < x < 1.
−
1
<
x
<
1.
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g
(
x
)
=
3
cot
(
x
2
)
−
csc
(
6
x
)
g(x)=3\cot(x^{2})-\csc(6x)
g
(
x
)
=
3
cot
(
x
2
)
−
csc
(
6
x
)
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What is the particular solution to the differential equation
d
y
d
x
=
(
1
−
y
)
2
x
+
1
\frac{dy}{dx}=\frac{(1-y)^{2}}{x+1}
d
x
d
y
=
x
+
1
(
1
−
y
)
2
with the initial condition
y
(
0
)
=
5
y(0)=5
y
(
0
)
=
5
?
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What is the particular solution to the differential equation
d
y
d
x
\frac{dy}{dx}
d
x
d
y
=
4
x
2
e
2
y
\frac{4}{x^{2}e^{2y}}
x
2
e
2
y
4
with the initial condition
y
(
1
)
=
0
y(1)=0
y
(
1
)
=
0
?
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Find the inverse of the function
\newline
f
(
x
)
=
3
x
2
−
27
,
x
≥
0
f(x)=3x^{2}-27,\,x \geq 0
f
(
x
)
=
3
x
2
−
27
,
x
≥
0
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Find the derivative of
\newline
f
(
x
)
=
−
4
x
(
3
x
4
)
.
f(x)=-4x(3^{x^{4}}).
f
(
x
)
=
−
4
x
(
3
x
4
)
.
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Integrate
1
1
+
x
2
\frac{1}{1+x^2}
1
+
x
2
1
for a limit
[
0
,
1
]
[0,1]
[
0
,
1
]
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What is the particular solution to the differential equation
\newline
d
y
d
x
=
(
x
−
3
)
y
\frac{dy}{dx}=(x-3)y
d
x
d
y
=
(
x
−
3
)
y
with the initial condition
\newline
y
(
0
)
=
2
y(0)=2
y
(
0
)
=
2
?
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