Resources
Testimonials
Plans
Sign in
Sign up
Resources
Testimonials
Plans
Home
Math Problems
Calculus
Find derivatives using the chain rule I
13
13
13
) Find the
\newline
f
(
x
)
=
2
x
−
3
x
3
+
3
x
f(x)=\frac{2 x-3}{x^{3}+3 x}
f
(
x
)
=
x
3
+
3
x
2
x
−
3
Get tutor help
y
=
(
x
−
1
)
2
y=(x-1)^{2}
y
=
(
x
−
1
)
2
Get tutor help
y
=
4
x
2
−
16
x
y=4x^{2}-16x
y
=
4
x
2
−
16
x
Get tutor help
lim
x
→
+
∞
x
ln
(
x
)
x
2
+
1
=
?
\lim _{x \rightarrow+\infty} \frac{\sqrt{x} \ln (x)}{x^{2}+1}=?
lim
x
→
+
∞
x
2
+
1
x
l
n
(
x
)
=
?
Get tutor help
Find the derivative of
y
=
tan
8
x
+
3
(
x
)
y=\tan ^{8 x+3}(x)
y
=
tan
8
x
+
3
(
x
)
.
Get tutor help
d)
y
=
e
7
x
y=e^{7 x}
y
=
e
7
x
Get tutor help
In the following equation, what is the value of
c
c
c
?
\newline
8
c
=
(
8
−
4
)
5
8^c = (8^{-4})^5
8
c
=
(
8
−
4
)
5
\newline
c
=
c =
c
=
____
Get tutor help
Find the derivative of
f
(
x
)
f(x)
f
(
x
)
. Where
f
(
x
)
=
e
(
x
+
1
)
f(x)=e^{(x+1)}
f
(
x
)
=
e
(
x
+
1
)
f
′
(
x
)
=
?
f'(x) = ?
f
′
(
x
)
=
?
Get tutor help
Pada gambar di atas, terdapat lingkaran dengan titik pusat
0
,
O
A
0, O \mathrm{~A}
0
,
O
A
dan
O
B
O \mathrm{~B}
O
B
adalah jari-jari lingkaran. Luas tembereng tersebut adalah ...
c
m
2
\mathrm{cm}^{2}
cm
2
.
Get tutor help
33
33
33
.
a
n
=
n
2
−
2
n
+
1
n
−
1
a_{n}=\frac{n^{2}-2 n+1}{n-1}
a
n
=
n
−
1
n
2
−
2
n
+
1
Get tutor help
find the area under the graph of each function over the given
\newline
y
=
x
2
+
x
+
1
[
2
,
3
]
y=x^{2}+x+1 \quad[2,3]
y
=
x
2
+
x
+
1
[
2
,
3
]
Get tutor help
2
2
2
.
g
(
x
)
=
(
x
−
6
)
2
g(x)=(x-6)^{2}
g
(
x
)
=
(
x
−
6
)
2
Get tutor help
Finding a Derivative of the function.
\newline
9
9
9
.
y
=
(
2
x
−
7
)
3
y=(2 x-7)^{3}
y
=
(
2
x
−
7
)
3
Get tutor help
Determine the asymptote of the function
\newline
f
(
x
)
=
(
−
9
)
7
∧
x
−
10
f(x)=(-9) 7^{\wedge} x-10
f
(
x
)
=
(
−
9
)
7
∧
x
−
10
Get tutor help
What's the answer?
\newline
−
6
2
+
36
=
?
-6^{2}+36=?
−
6
2
+
36
=
?
Get tutor help
slope for
y
=
−
6
x
+
4
y=-6x + 4
y
=
−
6
x
+
4
Get tutor help
Solve.
\newline
x
2
=
2
x
+
1
x^{2}=2 x+1
x
2
=
2
x
+
1
Get tutor help
y
=
(
1
−
x
)
2
(
2
x
+
3
)
y=(1-x)^2(2x+3)
y
=
(
1
−
x
)
2
(
2
x
+
3
)
Get tutor help
0
=
−
(
y
2
−
2
y
+
x
)
0=-\left(y^{2}-2 y+x\right)
0
=
−
(
y
2
−
2
y
+
x
)
Get tutor help
y
=
(
−
x
2
−
16
x
)
−
30
y=\left(-x^{2}-16 x\right)-30
y
=
(
−
x
2
−
16
x
)
−
30
Get tutor help
y
=
−
(
x
+
7
)
2
−
9
y=-(x+7)^2-9
y
=
−
(
x
+
7
)
2
−
9
Get tutor help
f
(
X
)
=
X
2
f(X) = X^2
f
(
X
)
=
X
2
f
(
X
)
=
?
f(X) = ?
f
(
X
)
=
?
Get tutor help
Find derivative
\newline
b)
f
(
x
)
=
(
x
4
+
3
x
2
−
1
)
(
x
3
−
5
)
f(x)=\left(x^{4}+3 x^{2}-1\right)\left(x^{3}-5\right)
f
(
x
)
=
(
x
4
+
3
x
2
−
1
)
(
x
3
−
5
)
Get tutor help
2
2
2
.)
y
=
x
−
6
y=x-6
y
=
x
−
6
.
Get tutor help
y
′
+
2
y
x
−
3
=
sin
4
x
(
x
−
3
)
y^{\prime}+2 \frac{y}{x-3}=\frac{\sin 4 x}{(x-3)}
y
′
+
2
x
−
3
y
=
(
x
−
3
)
s
i
n
4
x
Get tutor help
Solve the equation:
y
=
x
2
+
8
x
y=x^{2}+8x
y
=
x
2
+
8
x
Get tutor help
Find the inverse of
\newline
f
(
x
)
=
1
4
x
−
7
f(x)=\frac{1}{4x}-7
f
(
x
)
=
4
x
1
−
7
Get tutor help
a
2
−
121
a^{2}-121
a
2
−
121
Get tutor help
f
(
x
)
=
x
2
−
8
x
+
16
f(x)=x^{2}-8 x+16
f
(
x
)
=
x
2
−
8
x
+
16
Get tutor help
What is the discriminant of the quadratic equation
−
x
2
−
x
−
2
=
0
-x^{2}-x-2=0
−
x
2
−
x
−
2
=
0
?
\newline
−
9
-9
−
9
\newline
−
7
-7
−
7
\newline
9
9
9
\newline
7
7
7
Get tutor help
f
(
x
)
=
−
(
x
+
2
)
2
+
16
f(x) = -(x+2)^2 + 16
f
(
x
)
=
−
(
x
+
2
)
2
+
16
Get tutor help
A(x)=(
8
8
8
+
2
2
2
x)^
2
2
2
Get tutor help
graph
g
(
x
)
=
(
x
+
5
)
2
−
2
g(x)=(x+5)^2-2
g
(
x
)
=
(
x
+
5
)
2
−
2
Get tutor help
Find the inverse of each function.
\newline
f
(
x
)
=
−
2
−
3
x
2
f(x)=\frac{-2-3 x}{2}
f
(
x
)
=
2
−
2
−
3
x
Get tutor help
Find the derivative of
\newline
f
(
x
)
f(x)
f
(
x
)
.
\newline
f
(
x
)
=
e
x
x
f(x)=\frac{e^{x}}{x}
f
(
x
)
=
x
e
x
Get tutor help
Find the derivative
f
(
x
)
=
e
(
x
+
1
)
f(x)=e^{(x+1)}
f
(
x
)
=
e
(
x
+
1
)
Get tutor help
Evaluate.
\newline
−
3
3
⋅
(
1
375
)
1
3
=
\sqrt[3]{-3} \cdot\left(\frac{1}{375}\right)^{\frac{1}{3}}=
3
−
3
⋅
(
375
1
)
3
1
=
Get tutor help
Find the sum.
\newline
∑
k
=
1
38
(
6
k
−
105
)
=
\sum_{k=1}^{38}(6 k-105)=
k
=
1
∑
38
(
6
k
−
105
)
=
Get tutor help
Find
lim
x
→
3
f
(
x
)
\lim _{x \rightarrow 3} f(x)
lim
x
→
3
f
(
x
)
for
f
(
x
)
=
x
2
−
4
11
−
2
x
f(x)=\frac{x^{2}-4}{11-2 x}
f
(
x
)
=
11
−
2
x
x
2
−
4
.
Get tutor help
Find
lim
x
→
0
g
(
x
)
\lim _{x \rightarrow 0} g(x)
lim
x
→
0
g
(
x
)
for
g
(
x
)
=
4
x
2
−
15
4
x
+
3
g(x)=\frac{4 x^{2}-15}{4 x+3}
g
(
x
)
=
4
x
+
3
4
x
2
−
15
Get tutor help
∑
j
=
0
2
(
j
2
)
=
\sum_{j=0}^{2}\left(j^{2}\right)=
∑
j
=
0
2
(
j
2
)
=
Get tutor help
∑
k
=
1
3
(
k
−
4
)
=
\sum_{k=1}^{3}(k-4)=
∑
k
=
1
3
(
k
−
4
)
=
Get tutor help
∑
n
=
0
2
(
−
n
)
=
\sum_{n=0}^{2}(-n)=
∑
n
=
0
2
(
−
n
)
=
Get tutor help
∑
n
=
0
2
(
n
)
=
\sum_{n=0}^{2}(n)=
∑
n
=
0
2
(
n
)
=
Get tutor help
∑
k
=
0
1
(
2
−
k
)
=
\sum_{k=0}^{1}(2-k)=
∑
k
=
0
1
(
2
−
k
)
=
Get tutor help
∑
j
=
1
3
(
3
j
)
=
\sum_{j=1}^{3}(3 j)=
∑
j
=
1
3
(
3
j
)
=
Get tutor help
∑
x
=
1
3
(
4
x
)
=
\sum_{x=1}^{3}(4 x)=
∑
x
=
1
3
(
4
x
)
=
Get tutor help
∑
k
=
0
1
(
2
k
−
1
)
=
\sum_{k=0}^{1}(2 k-1)=
∑
k
=
0
1
(
2
k
−
1
)
=
Get tutor help
∑
y
=
1
2
(
1
−
y
)
=
\sum_{y=1}^{2}(1-y)=
∑
y
=
1
2
(
1
−
y
)
=
Get tutor help
∑
t
=
0
1
(
4
−
t
)
=
\sum_{t=0}^{1}(4-t)=
∑
t
=
0
1
(
4
−
t
)
=
Get tutor help
1
2
Next
