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Math Problems
Algebra 2
Simplify variable expressions using properties
Given
y
=
2
sec
4
(
x
)
y=2 \sec ^{4}(x)
y
=
2
sec
4
(
x
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
(
4
z
3
)
(
−
3
z
3
)
=
□
(4z^3)(-3z^3)= \Box
(
4
z
3
)
(
−
3
z
3
)
=
□
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(
4
z
3
)
(
−
3
z
3
)
=
□
(4z^3)(-3z^3)=\Box
(
4
z
3
)
(
−
3
z
3
)
=
□
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(
7
x
3
)
(
3
x
7
)
=
□
(7x^3)(3x^7)=\Box
(
7
x
3
)
(
3
x
7
)
=
□
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(
7
x
3
)
(
3
x
7
)
=
□
(7x^3)(3x^7)=\Box
(
7
x
3
)
(
3
x
7
)
=
□
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(
−
9
n
9
)
(
n
3
)
=
□
(-9n^9)(n^3)=\Box
(
−
9
n
9
)
(
n
3
)
=
□
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(
−
9
n
9
)
(
n
3
)
=
□
(-9n^9)(n^3)=\Box
(
−
9
n
9
)
(
n
3
)
=
□
Get tutor help
(
−
b
7
)
(
7
b
4
)
=
□
(-b^7)(7b^4)=\Box
(
−
b
7
)
(
7
b
4
)
=
□
Get tutor help
(
−
b
7
)
(
7
b
4
)
=
□
(-b^7)(7b^4)=\Box
(
−
b
7
)
(
7
b
4
)
=
□
Get tutor help
(
−
t
7
)
(
−
t
5
)
=
□
\left(-t^7\right)\left(-t^5\right)= \Box
(
−
t
7
)
(
−
t
5
)
=
□
Get tutor help
Given the function
y
=
4
(
x
2
+
6
x
−
6
)
4
y=4\left(x^{2}+6 x-6\right)^{4}
y
=
4
(
x
2
+
6
x
−
6
)
4
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
y
=
2
(
−
7
x
2
−
2
x
)
6
y=2\left(-7 x^{2}-2 x\right)^{6}
y
=
2
(
−
7
x
2
−
2
x
)
6
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
y
=
−
2
(
−
5
x
2
−
2
x
)
3
y=-2\left(-5 x^{2}-2 x\right)^{3}
y
=
−
2
(
−
5
x
2
−
2
x
)
3
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
y
=
−
3
(
4
x
2
−
9
x
)
4
y=-3\left(4 x^{2}-9 x\right)^{4}
y
=
−
3
(
4
x
2
−
9
x
)
4
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
4
x
+
y
2
+
2
−
y
3
=
−
2
x
2
4 x+y^{2}+2-y^{3}=-2 x^{2}
4
x
+
y
2
+
2
−
y
3
=
−
2
x
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
−
4
,
3
)
(-4,3)
(
−
4
,
3
)
\newline
Answer:
d
y
d
x
∣
(
−
4
,
3
)
=
\left.\frac{d y}{d x}\right|_{(-4,3)}=
d
x
d
y
∣
∣
(
−
4
,
3
)
=
Get tutor help
If
1
−
4
y
2
+
2
y
=
−
x
3
−
y
3
1-4 y^{2}+2 y=-x^{3}-y^{3}
1
−
4
y
2
+
2
y
=
−
x
3
−
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
3
,
−
2
)
(3,-2)
(
3
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
3
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(3,-2)}=
d
x
d
y
∣
∣
(
3
,
−
2
)
=
Get tutor help
If
−
y
2
=
x
2
−
x
+
3
y
3
-y^{2}=x^{2}-x+3 y^{3}
−
y
2
=
x
2
−
x
+
3
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
−
4
,
−
2
)
(-4,-2)
(
−
4
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
−
4
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(-4,-2)}=
d
x
d
y
∣
∣
(
−
4
,
−
2
)
=
Get tutor help
If
x
2
−
y
3
−
5
=
4
x
x^{2}-y^{3}-5=4 x
x
2
−
y
3
−
5
=
4
x
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
−
4
,
3
)
(-4,3)
(
−
4
,
3
)
.
\newline
Answer:
d
y
d
x
∣
(
−
4
,
3
)
=
\left.\frac{d y}{d x}\right|_{(-4,3)}=
d
x
d
y
∣
∣
(
−
4
,
3
)
=
Get tutor help
If
4
−
5
x
2
−
2
y
2
=
y
3
−
x
4-5 x^{2}-2 y^{2}=y^{3}-x
4
−
5
x
2
−
2
y
2
=
y
3
−
x
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
−
2
)
(1,-2)
(
1
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(1,-2)}=
d
x
d
y
∣
∣
(
1
,
−
2
)
=
Get tutor help
Given that
v
=
4
u
3
+
3
v=4 u^{3}+3
v
=
4
u
3
+
3
, find
d
d
u
(
3
u
2
−
2
sin
v
)
\frac{d}{d u}\left(3 u^{2}-2 \sin v\right)
d
u
d
(
3
u
2
−
2
sin
v
)
in terms of only
u
u
u
.
\newline
Answer:
Get tutor help
Given that
y
=
3
u
5
+
2
y=3 u^{5}+2
y
=
3
u
5
+
2
, find
d
d
u
(
2
y
3
−
2
sin
u
)
\frac{d}{d u}\left(2 y^{3}-2 \sin u\right)
d
u
d
(
2
y
3
−
2
sin
u
)
in terms of only
u
u
u
.
\newline
Answer:
Get tutor help
Given that
u
=
4
v
2
+
1
u=4 v^{2}+1
u
=
4
v
2
+
1
, find
d
d
v
(
5
v
5
−
4
sin
u
)
\frac{d}{d v}\left(5 v^{5}-4 \sin u\right)
d
v
d
(
5
v
5
−
4
sin
u
)
in terms of only
v
v
v
.
\newline
Answer:
Get tutor help
Evaluate. Write your answer as a whole number or as a simplified fraction.
\newline
(
2
7
)
/
(
2
3
)
=
(2^{7})/(2^{3})=
(
2
7
)
/
(
2
3
)
=
Get tutor help
9
−
3
÷
(
1
3
)
+
1
=
9-3\div\left(\frac{1}{3}\right)+1=
9
−
3
÷
(
3
1
)
+
1
=
Get tutor help
8
:
2
(
2
+
2
)
=
8:2(2+2)=
8
:
2
(
2
+
2
)
=
Get tutor help
Given the function
y
=
2
+
3
x
3
−
2
x
2
y=\frac{2+3 x}{3-2 x^{2}}
y
=
3
−
2
x
2
2
+
3
x
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in simplified form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
f
(
x
)
=
1
−
x
3
x
3
−
4
f(x)=\frac{1-x}{3 x^{3}-4}
f
(
x
)
=
3
x
3
−
4
1
−
x
, find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
in simplified form.
\newline
Answer:
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Get tutor help
Given the function
f
(
x
)
=
x
3
−
2
2
x
−
3
f(x)=\frac{x^{3}-2}{2 x-3}
f
(
x
)
=
2
x
−
3
x
3
−
2
, find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
in simplified form.
\newline
Answer:
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Get tutor help
Given the function
y
=
1
+
x
4
5
x
−
1
y=\frac{1+x^{4}}{5 x-1}
y
=
5
x
−
1
1
+
x
4
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in simplified form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
y
=
3
x
+
2
2
−
x
2
y=\frac{3 x+2}{2-x^{2}}
y
=
2
−
x
2
3
x
+
2
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in simplified form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
f
(
x
)
=
x
1
−
2
x
4
f(x)=\frac{x}{1-2 x^{4}}
f
(
x
)
=
1
−
2
x
4
x
, find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
in simplified form.
\newline
Answer:
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Get tutor help
Find
d
d
x
(
−
cos
(
−
x
+
5
)
)
\frac{d}{d x}(-\cos (-x+5))
d
x
d
(
−
cos
(
−
x
+
5
))
\newline
Answer:
Get tutor help
Find
d
d
x
(
7
cos
4
x
)
\frac{d}{d x}(7 \cos 4 x)
d
x
d
(
7
cos
4
x
)
\newline
Answer:
Get tutor help
Find
d
d
x
(
sin
(
−
2
x
+
7
)
)
\frac{d}{d x}(\sin (-2 x+7))
d
x
d
(
sin
(
−
2
x
+
7
))
\newline
Answer:
Get tutor help
Find
d
d
x
(
9
cos
(
−
x
+
1
)
)
\frac{d}{d x}(9 \cos (-x+1))
d
x
d
(
9
cos
(
−
x
+
1
))
\newline
Answer:
Get tutor help
Find
d
d
x
(
−
cos
x
+
5
)
\frac{d}{d x}(-\cos x+5)
d
x
d
(
−
cos
x
+
5
)
\newline
Answer:
Get tutor help
Given the function
y
=
(
2
x
3
−
1
)
(
3
x
2
−
3
+
x
)
y=\left(2 x^{3}-1\right)\left(3 x^{2}-3+x\right)
y
=
(
2
x
3
−
1
)
(
3
x
2
−
3
+
x
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
y
=
(
−
1
+
2
x
)
(
−
7
x
−
3
−
3
+
3
x
)
y=(-1+2 x)\left(-7 x^{-3}-3+3 x\right)
y
=
(
−
1
+
2
x
)
(
−
7
x
−
3
−
3
+
3
x
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Simplify
ln
(
e
4
3
)
\ln \left(\sqrt[3]{e^{4}}\right)
ln
(
3
e
4
)
\newline
Answer:
Get tutor help
Simplify
ln
(
e
2
)
\ln \left(e^{2}\right)
ln
(
e
2
)
\newline
Answer:
Get tutor help
Simplify
e
−
ln
6
e^{-\ln 6}
e
−
l
n
6
\newline
Answer:
Get tutor help
Simplify
e
−
ln
3
e^{-\ln 3}
e
−
l
n
3
\newline
Answer:
Get tutor help
Let
h
(
x
)
=
x
5
h(x)=\sqrt{x^{5}}
h
(
x
)
=
x
5
.
\newline
h
′
(
4
)
=
h^{\prime}(4)=
h
′
(
4
)
=
Get tutor help
f
′
(
x
)
=
6
x
5
+
7
and
f
(
−
2
)
=
30.
\begin{array}{l}f^{\prime}(x)=6 x^{5}+7 \text { and } f(-2)=30 .\end{array}
f
′
(
x
)
=
6
x
5
+
7
and
f
(
−
2
)
=
30.
\newline
f
(
1
)
=
f(1)=
f
(
1
)
=
Get tutor help
f
′
(
x
)
=
16
x
3
+
6
x
−
7
and
f
(
5
)
=
2500.
\begin{array}{l}f^{\prime}(x)=16 x^{3}+6 x-7 \text { and } f(5)=2500 .\end{array}
f
′
(
x
)
=
16
x
3
+
6
x
−
7
and
f
(
5
)
=
2500.
\newline
f
(
−
1
)
=
f(-1)=
f
(
−
1
)
=
Get tutor help
d
d
x
(
1
x
9
)
=
\frac{d}{d x}\left(\frac{1}{x^{9}}\right)=
d
x
d
(
x
9
1
)
=
Get tutor help
d
d
x
(
1
x
8
)
=
\frac{d}{d x}\left(\frac{1}{x^{8}}\right)=
d
x
d
(
x
8
1
)
=
Get tutor help
d
d
x
(
x
−
9
)
=
\frac{d}{d x}\left(x^{-9}\right)=
d
x
d
(
x
−
9
)
=
Get tutor help
d
d
x
(
x
5
3
)
=
\frac{d}{d x}\left(x^{\frac{5}{3}}\right)=
d
x
d
(
x
3
5
)
=
Get tutor help
d
d
x
(
x
−
4
)
=
\frac{d}{d x}\left(x^{-4}\right)=
d
x
d
(
x
−
4
)
=
Get tutor help
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