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Math Problems
Calculus
Find derivatives of using multiple formulae
Let
y
=
e
x
x
y=\frac{e^{x}}{x}
y
=
x
e
x
.
\newline
Find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Choose
1
1
1
answer:
\newline
(A)
e
x
e^{x}
e
x
\newline
(B)
e
x
−
1
x
2
\frac{e^{x}-1}{x^{2}}
x
2
e
x
−
1
\newline
(C)
e
x
−
1
x
2
\frac{e^{x-1}}{x^{2}}
x
2
e
x
−
1
\newline
(D)
e
x
(
x
−
1
)
x
2
\frac{e^{x}(x-1)}{x^{2}}
x
2
e
x
(
x
−
1
)
Get tutor help
Let
g
(
x
)
=
cos
(
x
)
sin
(
x
)
g(x)=\frac{\cos (x)}{\sin (x)}
g
(
x
)
=
s
i
n
(
x
)
c
o
s
(
x
)
.
\newline
Find
g
′
(
x
)
g^{\prime}(x)
g
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1
sin
2
(
x
)
-\frac{1}{\sin ^{2}(x)}
−
s
i
n
2
(
x
)
1
\newline
(B)
sin
2
(
x
)
−
cos
2
(
x
)
sin
2
(
x
)
\frac{\sin ^{2}(x)-\cos ^{2}(x)}{\sin ^{2}(x)}
s
i
n
2
(
x
)
s
i
n
2
(
x
)
−
c
o
s
2
(
x
)
\newline
(C)
1
sin
2
(
x
)
\frac{1}{\sin ^{2}(x)}
s
i
n
2
(
x
)
1
\newline
(D)
cos
2
(
x
)
−
sin
2
(
x
)
sin
2
(
x
)
\frac{\cos ^{2}(x)-\sin ^{2}(x)}{\sin ^{2}(x)}
s
i
n
2
(
x
)
c
o
s
2
(
x
)
−
s
i
n
2
(
x
)
Get tutor help
Let
g
(
x
)
=
sin
(
x
)
e
x
g(x)=\frac{\sin (x)}{e^{x}}
g
(
x
)
=
e
x
s
i
n
(
x
)
.
\newline
Find
g
′
(
x
)
g^{\prime}(x)
g
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
cos
(
x
)
−
e
x
\cos (x)-e^{x}
cos
(
x
)
−
e
x
\newline
(B)
e
x
(
cos
(
x
)
−
sin
(
x
)
)
e^{x}(\cos (x)-\sin (x))
e
x
(
cos
(
x
)
−
sin
(
x
))
\newline
(C)
cos
(
x
)
e
x
\frac{\cos (x)}{e^{x}}
e
x
c
o
s
(
x
)
\newline
(D)
cos
(
x
)
−
sin
(
x
)
e
x
\frac{\cos (x)-\sin (x)}{e^{x}}
e
x
c
o
s
(
x
)
−
s
i
n
(
x
)
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
8
x
3
y
−
8
x
y
\frac{d y}{d x}=8 x^{3} y-8 x y
d
x
d
y
=
8
x
3
y
−
8
x
y
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
C
e
2
x
4
−
4
x
2
y=C e^{2 x^{4}-4 x^{2}}
y
=
C
e
2
x
4
−
4
x
2
\newline
(B)
y
=
e
2
x
4
−
4
x
2
+
C
y=e^{2 x^{4}-4 x^{2}}+C
y
=
e
2
x
4
−
4
x
2
+
C
\newline
(C)
y
=
2
x
4
−
4
x
2
+
C
y=2 x^{4}-4 x^{2}+C
y
=
2
x
4
−
4
x
2
+
C
\newline
(D)
y
=
C
(
2
x
4
−
4
x
2
)
y=C\left(2 x^{4}-4 x^{2}\right)
y
=
C
(
2
x
4
−
4
x
2
)
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
x
e
−
y
+
10
e
−
y
\frac{d y}{d x}=x e^{-y}+10 e^{-y}
d
x
d
y
=
x
e
−
y
+
10
e
−
y
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
x
2
2
+
10
x
+
C
y=\frac{x^{2}}{2}+10 x+C
y
=
2
x
2
+
10
x
+
C
\newline
(B)
y
=
−
ln
(
−
x
2
2
−
10
x
+
C
)
y=-\ln \left(-\frac{x^{2}}{2}-10 x+C\right)
y
=
−
ln
(
−
2
x
2
−
10
x
+
C
)
\newline
(C)
y
=
ln
(
x
2
2
+
10
x
+
C
)
y=\ln \left(\frac{x^{2}}{2}+10 x+C\right)
y
=
ln
(
2
x
2
+
10
x
+
C
)
\newline
(D)
y
=
−
ln
(
−
x
2
2
−
10
x
)
+
C
y=-\ln \left(-\frac{x^{2}}{2}-10 x\right)+C
y
=
−
ln
(
−
2
x
2
−
10
x
)
+
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
y
−
1
x
−
3
\frac{d y}{d x}=y^{-1} x^{-3}
d
x
d
y
=
y
−
1
x
−
3
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
±
x
4
2
+
C
y= \pm \sqrt{\frac{x^{4}}{2}+C}
y
=
±
2
x
4
+
C
\newline
(B)
y
=
±
x
4
2
+
C
y= \pm \sqrt{\frac{x^{4}}{2}}+C
y
=
±
2
x
4
+
C
\newline
(C)
y
=
±
−
1
x
2
+
C
y= \pm \sqrt{-\frac{1}{x^{2}}}+C
y
=
±
−
x
2
1
+
C
\newline
(D)
y
=
±
−
1
x
2
+
C
y= \pm \sqrt{-\frac{1}{x^{2}}+C}
y
=
±
−
x
2
1
+
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
−
e
x
y
2
10
\frac{d y}{d x}=-\frac{e^{x} y^{2}}{10}
d
x
d
y
=
−
10
e
x
y
2
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
10
e
x
+
C
y=\frac{10}{e^{x}}+C
y
=
e
x
10
+
C
\newline
(B)
y
=
10
e
x
+
C
y=\frac{10}{e^{x}+C}
y
=
e
x
+
C
10
\newline
(C)
y
=
10
C
e
x
y=\frac{10}{C e^{x}}
y
=
C
e
x
10
\newline
(D)
y
=
10
+
C
e
x
y=\frac{10+C}{e^{x}}
y
=
e
x
10
+
C
Get tutor help
Marcus tried to solve the differential equation
d
y
d
x
=
1
3
x
2
y
2
\frac{d y}{d x}=\frac{1}{3 x^{2} y^{2}}
d
x
d
y
=
3
x
2
y
2
1
. This is his work:
\newline
d
y
d
x
=
1
3
x
2
y
2
\frac{d y}{d x}=\frac{1}{3 x^{2} y^{2}}
d
x
d
y
=
3
x
2
y
2
1
\newline
Step
1
1
1
:
∫
3
y
2
d
y
=
∫
1
x
2
d
x
\quad \int 3 y^{2} d y=\int \frac{1}{x^{2}} d x
∫
3
y
2
d
y
=
∫
x
2
1
d
x
\newline
Step
2
2
2
:
y
3
=
−
1
x
+
C
\quad y^{3}=-\frac{1}{x}+C
y
3
=
−
x
1
+
C
\newline
Step
3
3
3
:
y
=
±
−
1
x
+
C
3
\quad y= \pm \sqrt[3]{-\frac{1}{x}+C}
y
=
±
3
−
x
1
+
C
\newline
Is Marcus's work correct? If not, what is his mistake?
\newline
Choose
1
1
1
answer:
\newline
(A) Marcus's work is correct.
\newline
(B) Step
2
2
2
is incorrect. Marcus didn't integrate
3
y
2
3 y^{2}
3
y
2
correctly.
\newline
(C) Step
3
3
3
is incorrect. Marcus forgot to add a constant to the right-hand side at the end of the process.
\newline
(D) Step
3
3
3
is incorrect. The right-hand side of the equation should be
−
1
x
+
C
3
\sqrt[3]{-\frac{1}{x}+C}
3
−
x
1
+
C
.
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
−
1
3
x
3
\frac{d y}{d x}=-\frac{1}{3} x^{3}
d
x
d
y
=
−
3
1
x
3
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
−
x
4
12
+
C
y=-\frac{x^{4}}{12}+C
y
=
−
12
x
4
+
C
\newline
(B)
y
=
−
x
2
+
C
y=-x^{2}+C
y
=
−
x
2
+
C
\newline
(c)
y
=
−
x
4
12
+
C
y=-\frac{x^{4}}{12+C}
y
=
−
12
+
C
x
4
\newline
(D)
y
=
−
x
2
+
C
y=-x^{2+C}
y
=
−
x
2
+
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
−
1
4
e
x
y
−
2
\frac{d y}{d x}=-\frac{1}{4} e^{x} y^{-2}
d
x
d
y
=
−
4
1
e
x
y
−
2
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
±
−
3
e
x
4
+
C
3
y= \pm \sqrt[3]{-\frac{3 e^{x}}{4}+C}
y
=
±
3
−
4
3
e
x
+
C
\newline
(B)
y
=
−
3
e
x
4
+
C
3
y=\sqrt[3]{-\frac{3 e^{x}}{4}+C}
y
=
3
−
4
3
e
x
+
C
\newline
(C)
y
=
±
−
3
e
x
4
3
+
C
y= \pm \sqrt[3]{-\frac{3 e^{x}}{4}}+C
y
=
±
3
−
4
3
e
x
+
C
\newline
(D)
y
=
−
3
e
x
4
3
+
C
y=\sqrt[3]{-\frac{3 e^{x}}{4}}+C
y
=
3
−
4
3
e
x
+
C
Get tutor help
Let
f
(
x
)
=
3
x
2
+
1
x
+
2
f(x)=\frac{3 x^{2}+1}{x+2}
f
(
x
)
=
x
+
2
3
x
2
+
1
.
\newline
Find
f
′
(
−
3
)
f^{\prime}(-3)
f
′
(
−
3
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
10
\mathbf{1 0}
10
\newline
(B)
−
10
-10
−
10
\newline
(C)
−
28
-28
−
28
\newline
(D)
28
28
28
Get tutor help
What is the value of
d
d
x
(
x
2
−
2
x
+
3
x
+
1
)
\frac{d}{d x}\left(\frac{x^{2}-2 x+3}{x+1}\right)
d
x
d
(
x
+
1
x
2
−
2
x
+
3
)
at
x
=
1
x=1
x
=
1
?
\newline
Choose
1
1
1
answer:
\newline
(A)
1
1
1
\newline
(B)
−
1
2
-\frac{1}{2}
−
2
1
\newline
(C)
−
2
-2
−
2
\newline
(D)
−
1
-1
−
1
Get tutor help
Find
d
d
x
(
e
x
x
)
\frac{d}{d x}\left(\frac{e^{x}}{\sqrt{x}}\right)
d
x
d
(
x
e
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
e
x
−
1
2
x
e^{x}-\frac{1}{2 \sqrt{x}}
e
x
−
2
x
1
\newline
(B)
e
x
(
x
−
1
2
x
)
x
\frac{e^{x}\left(\sqrt{x}-\frac{1}{2 \sqrt{x}}\right)}{x}
x
e
x
(
x
−
2
x
1
)
\newline
(C)
2
x
e
x
2 \sqrt{x} e^{x}
2
x
e
x
\newline
(D)
e
x
(
x
−
1
2
x
)
e^{x}\left(\sqrt{x}-\frac{1}{2 \sqrt{x}}\right)
e
x
(
x
−
2
x
1
)
Get tutor help
Let
f
(
x
)
=
x
2
ln
(
x
)
f(x)=\frac{x^{2}}{\ln (x)}
f
(
x
)
=
l
n
(
x
)
x
2
.
\newline
Find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
2
x
ln
(
x
)
−
x
(
ln
(
x
)
)
2
\frac{2 x \ln (x)-x}{(\ln (x))^{2}}
(
l
n
(
x
)
)
2
2
x
l
n
(
x
)
−
x
\newline
(B)
2
x
ln
(
x
)
+
x
2 x \ln (x)+x
2
x
ln
(
x
)
+
x
\newline
(C)
2
x
−
1
x
2 x-\frac{1}{x}
2
x
−
x
1
\newline
(D)
2
x
2
2 x^{2}
2
x
2
Get tutor help
Let
h
(
x
)
=
cos
(
x
)
ln
(
x
)
h(x)=\frac{\cos (x)}{\ln (x)}
h
(
x
)
=
l
n
(
x
)
c
o
s
(
x
)
.
\newline
Find
h
′
(
x
)
h^{\prime}(x)
h
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
x
sin
(
x
)
ln
(
x
)
−
cos
(
x
)
x
(
ln
(
x
)
)
2
\frac{-x \sin (x) \ln (x)-\cos (x)}{x(\ln (x))^{2}}
x
(
l
n
(
x
)
)
2
−
x
s
i
n
(
x
)
l
n
(
x
)
−
c
o
s
(
x
)
\newline
(B)
−
x
sin
(
x
)
-x \sin (x)
−
x
sin
(
x
)
\newline
(C)
ln
(
x
)
sin
(
x
)
−
cos
(
x
)
x
ln
(
x
)
\frac{\ln (x) \sin (x)-\cos (x)}{x \ln (x)}
x
l
n
(
x
)
l
n
(
x
)
s
i
n
(
x
)
−
c
o
s
(
x
)
\newline
(D)
sin
(
x
)
−
1
x
\sin (x)-\frac{1}{x}
sin
(
x
)
−
x
1
Get tutor help
Let
y
=
e
x
x
y=\frac{e^{x}}{x}
y
=
x
e
x
.
\newline
Find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Choose
1
1
1
answer:
\newline
(A)
e
x
−
1
x
2
\frac{e^{x-1}}{x^{2}}
x
2
e
x
−
1
\newline
(B)
e
x
(
x
−
1
)
x
2
\frac{e^{x}(x-1)}{x^{2}}
x
2
e
x
(
x
−
1
)
\newline
(C)
e
x
e^{x}
e
x
\newline
(D)
e
x
−
1
x
2
\frac{e^{x}-1}{x^{2}}
x
2
e
x
−
1
Get tutor help
Let
g
(
x
)
=
cos
(
x
)
sin
(
x
)
g(x)=\frac{\cos (x)}{\sin (x)}
g
(
x
)
=
s
i
n
(
x
)
c
o
s
(
x
)
.
\newline
Find
g
′
(
x
)
g^{\prime}(x)
g
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
sin
2
(
x
)
−
cos
2
(
x
)
sin
2
(
x
)
\frac{\sin ^{2}(x)-\cos ^{2}(x)}{\sin ^{2}(x)}
s
i
n
2
(
x
)
s
i
n
2
(
x
)
−
c
o
s
2
(
x
)
\newline
(B)
1
sin
2
(
x
)
\frac{1}{\sin ^{2}(x)}
s
i
n
2
(
x
)
1
\newline
(C)
−
1
sin
2
(
x
)
-\frac{1}{\sin ^{2}(x)}
−
s
i
n
2
(
x
)
1
\newline
(D)
cos
2
(
x
)
−
sin
2
(
x
)
sin
2
(
x
)
\frac{\cos ^{2}(x)-\sin ^{2}(x)}{\sin ^{2}(x)}
s
i
n
2
(
x
)
c
o
s
2
(
x
)
−
s
i
n
2
(
x
)
Get tutor help
Let
g
(
x
)
=
sin
(
x
)
e
x
g(x)=\frac{\sin (x)}{e^{x}}
g
(
x
)
=
e
x
s
i
n
(
x
)
.
\newline
Find
g
′
(
x
)
g^{\prime}(x)
g
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
cos
(
x
)
e
x
\frac{\cos (x)}{e^{x}}
e
x
c
o
s
(
x
)
\newline
(B)
cos
(
x
)
−
e
x
\cos (x)-e^{x}
cos
(
x
)
−
e
x
\newline
(C)
e
x
(
cos
(
x
)
−
sin
(
x
)
)
e^{x}(\cos (x)-\sin (x))
e
x
(
cos
(
x
)
−
sin
(
x
))
\newline
(D)
cos
(
x
)
−
sin
(
x
)
e
x
\frac{\cos (x)-\sin (x)}{e^{x}}
e
x
c
o
s
(
x
)
−
s
i
n
(
x
)
Get tutor help
Jaxton tried to solve the differential equation
d
y
d
x
=
(
sec
(
x
)
y
)
2
\frac{d y}{d x}=\left(\frac{\sec (x)}{y}\right)^{2}
d
x
d
y
=
(
y
s
e
c
(
x
)
)
2
. This is his work:
\newline
d
y
d
x
=
(
sec
(
x
)
y
)
2
\frac{d y}{d x}=\left(\frac{\sec (x)}{y}\right)^{2}
d
x
d
y
=
(
y
sec
(
x
)
)
2
\newline
Step
1
1
1
:
d
y
d
x
=
sec
2
(
x
)
y
2
\quad \frac{d y}{d x}=\frac{\sec ^{2}(x)}{y^{2}}
d
x
d
y
=
y
2
s
e
c
2
(
x
)
\newline
Step
2
2
2
:
∫
y
2
d
y
=
∫
sec
2
(
x
)
d
x
\quad \int y^{2} d y=\int \sec ^{2}(x) d x
∫
y
2
d
y
=
∫
sec
2
(
x
)
d
x
\newline
Step
3
3
3
:
y
3
3
=
tan
(
x
)
+
C
1
\quad \frac{y^{3}}{3}=\tan (x)+C_{1}
3
y
3
=
tan
(
x
)
+
C
1
\newline
Step
4
4
4
:
y
3
=
3
tan
(
x
)
+
C
\quad y^{3}=3 \tan (x)+C
y
3
=
3
tan
(
x
)
+
C
\newline
Step
5
5
5
:
y
=
3
tan
(
x
)
+
C
3
\quad y=\sqrt[3]{3 \tan (x)+C}
y
=
3
3
tan
(
x
)
+
C
\newline
Is Jaxton's work correct? If not, what is his mistake?
\newline
Choose
1
1
1
answer:
\newline
(A) Jaxton's work is correct.
\newline
(B) Step
1
1
1
is incorrect. The separation of variables was done incorrectly.
\newline
(C) Step
3
3
3
is incorrect. Jaxton didn't integrate
sec
2
(
x
)
\sec ^{2}(x)
sec
2
(
x
)
correctly.
\newline
(D) Step
5
5
5
is incorrect. The right-hand side of the equation should be
±
tan
(
x
)
+
C
3
\pm \sqrt[3]{\tan (x)+C}
±
3
tan
(
x
)
+
C
.
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
−
x
+
6
7
y
2
\frac{d y}{d x}=-\frac{x+6}{7 y^{2}}
d
x
d
y
=
−
7
y
2
x
+
6
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
±
−
x
+
6
7
+
C
y= \pm \sqrt{-\frac{x+6}{7}+C}
y
=
±
−
7
x
+
6
+
C
\newline
(B)
y
=
C
−
x
+
6
7
y=C \sqrt{-\frac{x+6}{7}}
y
=
C
−
7
x
+
6
\newline
(C)
y
=
C
−
3
x
2
14
−
18
x
7
3
y=C \sqrt[3]{-\frac{3 x^{2}}{14}-\frac{18 x}{7}}
y
=
C
3
−
14
3
x
2
−
7
18
x
\newline
(D)
y
=
−
3
x
2
14
−
18
x
7
+
C
3
y=\sqrt[3]{-\frac{3 x^{2}}{14}-\frac{18 x}{7}+C}
y
=
3
−
14
3
x
2
−
7
18
x
+
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
x
7
cos
(
y
)
\frac{d y}{d x}=\frac{x}{7 \cos (y)}
d
x
d
y
=
7
cos
(
y
)
x
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
cos
(
x
)
+
x
sin
(
x
)
7
cos
2
(
x
)
+
C
y=\frac{\cos (x)+x \sin (x)}{7 \cos ^{2}(x)}+C
y
=
7
c
o
s
2
(
x
)
c
o
s
(
x
)
+
x
s
i
n
(
x
)
+
C
\newline
(B)
y
=
arcsin
(
x
2
14
+
C
)
y=\arcsin \left(\frac{x^{2}}{14}+C\right)
y
=
arcsin
(
14
x
2
+
C
)
\newline
(C)
y
=
cos
(
x
)
+
x
sin
(
x
)
+
C
7
cos
2
(
x
)
y=\frac{\cos (x)+x \sin (x)+C}{7 \cos ^{2}(x)}
y
=
7
c
o
s
2
(
x
)
c
o
s
(
x
)
+
x
s
i
n
(
x
)
+
C
\newline
(D)
y
=
arcsin
(
x
2
14
)
+
C
y=\arcsin \left(\frac{x^{2}}{14}\right)+C
y
=
arcsin
(
14
x
2
)
+
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
4
x
3
y
−
6
x
2
y
\frac{d y}{d x}=4 x^{3} y-6 x^{2} y
d
x
d
y
=
4
x
3
y
−
6
x
2
y
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
e
x
4
−
2
x
3
+
C
y=e^{x^{4}-2 x^{3}}+C
y
=
e
x
4
−
2
x
3
+
C
\newline
(B)
y
=
±
x
4
−
2
x
3
+
C
y= \pm \sqrt{x^{4}-2 x^{3}+C}
y
=
±
x
4
−
2
x
3
+
C
\newline
(C)
y
=
C
e
x
4
−
2
x
3
y=C e^{x^{4}-2 x^{3}}
y
=
C
e
x
4
−
2
x
3
\newline
(D)
y
=
C
x
4
−
2
x
3
y=C \sqrt{x^{4}-2} x^{3}
y
=
C
x
4
−
2
x
3
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
x
2
10
y
−
2
x
5
y
\frac{d y}{d x}=\frac{x^{2}}{10 y}-\frac{2 x}{5 y}
d
x
d
y
=
10
y
x
2
−
5
y
2
x
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
±
x
3
15
−
2
x
2
5
+
C
y= \pm \sqrt{\frac{x^{3}}{15}-\frac{2 x^{2}}{5}+C}
y
=
±
15
x
3
−
5
2
x
2
+
C
\newline
(B)
y
=
±
x
3
15
−
2
x
2
5
+
C
y= \pm \sqrt{\frac{x^{3}}{15}-\frac{2 x^{2}}{5}}+C
y
=
±
15
x
3
−
5
2
x
2
+
C
\newline
(C)
y
=
C
e
10
x
3
3
−
20
x
2
y=C e^{\frac{10 x^{3}}{3}-20 x^{2}}
y
=
C
e
3
10
x
3
−
20
x
2
\newline
(D)
y
=
±
e
10
x
3
3
−
20
x
2
+
C
y= \pm e^{\frac{10 x^{3}}{3}-20 x^{2}}+C
y
=
±
e
3
10
x
3
−
20
x
2
+
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
4
y
e
x
\frac{d y}{d x}=4 y e^{x}
d
x
d
y
=
4
y
e
x
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
e
C
e
x
y=e^{C e^{x}}
y
=
e
C
e
x
\newline
(B)
y
=
e
e
x
+
C
y=e^{e^{x}+C}
y
=
e
e
x
+
C
\newline
(C)
y
=
e
4
e
x
+
C
y=e^{4 e^{x}}+C
y
=
e
4
e
x
+
C
\newline
(D)
y
=
C
e
4
e
x
y=C e^{4 e^{x}}
y
=
C
e
4
e
x
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
x
2
9
+
7
9
\frac{d y}{d x}=\frac{x^{2}}{9}+\frac{7}{9}
d
x
d
y
=
9
x
2
+
9
7
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
2
x
9
+
C
y=\frac{2 x}{9}+C
y
=
9
2
x
+
C
\newline
(B)
y
=
x
18
+
C
y=\frac{x}{18}+C
y
=
18
x
+
C
\newline
(C)
y
=
x
3
3
+
7
x
9
+
C
y=\frac{x^{3}}{3}+\frac{7 x}{9}+C
y
=
3
x
3
+
9
7
x
+
C
\newline
(D)
y
=
x
3
27
+
7
x
9
+
C
y=\frac{x^{3}}{27}+\frac{7 x}{9}+C
y
=
27
x
3
+
9
7
x
+
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
x
sin
(
y
)
−
9
sin
(
y
)
\frac{d y}{d x}=\frac{x}{\sin (y)}-\frac{9}{\sin (y)}
d
x
d
y
=
sin
(
y
)
x
−
sin
(
y
)
9
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
arccos
(
−
x
2
2
+
9
x
+
C
)
y=\arccos \left(-\frac{x^{2}}{2}+9 x+C\right)
y
=
arccos
(
−
2
x
2
+
9
x
+
C
)
\newline
(B)
y
=
arccos
(
−
x
2
2
+
9
x
)
+
C
y=\arccos \left(-\frac{x^{2}}{2}+9 x\right)+C
y
=
arccos
(
−
2
x
2
+
9
x
)
+
C
\newline
(C)
y
=
2
cos
(
−
x
2
+
18
x
+
C
)
y=\frac{2}{\cos \left(-x^{2}+18 x+C\right)}
y
=
c
o
s
(
−
x
2
+
18
x
+
C
)
2
\newline
(D)
y
=
2
C
cos
(
−
x
2
+
18
x
)
y=\frac{2 C}{\cos \left(-x^{2}+18 x\right)}
y
=
c
o
s
(
−
x
2
+
18
x
)
2
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
x
2
6
e
y
\frac{d y}{d x}=\frac{x^{2}}{6 e^{y}}
d
x
d
y
=
6
e
y
x
2
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
±
−
2
e
−
x
+
C
+
10
y= \pm \sqrt{-2 e^{-x}+C}+10
y
=
±
−
2
e
−
x
+
C
+
10
\newline
(B)
y
=
±
−
2
e
−
x
+
C
y= \pm \sqrt{-2 e^{-x}}+C
y
=
±
−
2
e
−
x
+
C
\newline
(C)
y
=
ln
(
x
3
18
+
C
)
y=\ln \left(\frac{x^{3}}{18}+C\right)
y
=
ln
(
18
x
3
+
C
)
\newline
(D)
y
=
ln
(
x
3
18
)
+
C
y=\ln \left(\frac{x^{3}}{18}\right)+C
y
=
ln
(
18
x
3
)
+
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
(
x
y
)
2
8
+
y
2
\frac{d y}{d x}=\frac{(x y)^{2}}{8}+y^{2}
d
x
d
y
=
8
(
x
y
)
2
+
y
2
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
−
x
2
+
16
16
+
C
y=-\frac{x^{2}+16}{16+C}
y
=
−
16
+
C
x
2
+
16
\newline
(B)
y
=
−
16
x
2
+
16
+
C
y=-\frac{16}{x^{2}+16+C}
y
=
−
x
2
+
16
+
C
16
\newline
(C)
y
=
−
x
3
+
24
x
24
+
C
y=-\frac{x^{3}+24 x}{24+C}
y
=
−
24
+
C
x
3
+
24
x
\newline
(D)
y
=
−
24
x
3
+
24
x
+
C
y=-\frac{24}{x^{3}+24 x+C}
y
=
−
x
3
+
24
x
+
C
24
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
−
10
x
4
y
\frac{d y}{d x}=-10 x^{4} y
d
x
d
y
=
−
10
x
4
y
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
−
2
(
x
+
C
)
5
y=-2(x+C)^{5}
y
=
−
2
(
x
+
C
)
5
\newline
(B)
y
=
C
e
−
2
x
5
y=C e^{-2 x^{5}}
y
=
C
e
−
2
x
5
\newline
(C)
y
=
−
2
x
5
+
C
y=-2 x^{5}+C
y
=
−
2
x
5
+
C
\newline
(D)
y
=
e
−
2
x
5
+
C
y=e^{-2 x^{5}}+C
y
=
e
−
2
x
5
+
C
Get tutor help
Solve the equation.
\newline
d
y
d
x
=
4
x
3
cos
(
y
)
\frac{d y}{d x}=\frac{4 x^{3}}{\cos (y)}
d
x
d
y
=
cos
(
y
)
4
x
3
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
arccos
(
4
x
3
)
+
C
y=\arccos \left(4 x^{3}\right)+C
y
=
arccos
(
4
x
3
)
+
C
\newline
(B)
y
=
arccos
(
4
x
3
+
C
)
y=\arccos \left(4 x^{3}+C\right)
y
=
arccos
(
4
x
3
+
C
)
\newline
(C)
y
=
arcsin
(
x
4
+
C
)
y=\arcsin \left(x^{4}+C\right)
y
=
arcsin
(
x
4
+
C
)
\newline
(D)
y
=
arcsin
(
x
4
)
+
C
y=\arcsin \left(x^{4}\right)+C
y
=
arcsin
(
x
4
)
+
C
Get tutor help
The base of a solid is the region enclosed by the graphs of
y
=
2
+
x
2
y=2+\frac{x}{2}
y
=
2
+
2
x
and
y
=
2
x
2
y=2^{\frac{x}{2}}
y
=
2
2
x
, between the
y
y
y
-axis and
x
=
4
x=4
x
=
4
.
\newline
Cross sections of the solid perpendicular to the
x
x
x
-axis are rectangles whose height is
4
−
x
4-x
4
−
x
.
\newline
Which one of the definite integrals gives the volume of the solid?
\newline
Choose
1
1
1
answer:
\newline
(A)
∫
0
4
[
2
+
x
2
−
2
x
2
]
(
4
−
x
)
d
x
\int_{0}^{4}\left[2+\frac{x}{2}-2^{\frac{x}{2}}\right](4-x) d x
∫
0
4
[
2
+
2
x
−
2
2
x
]
(
4
−
x
)
d
x
\newline
(B)
∫
0
4
[
2
x
2
+
x
2
+
2
]
(
4
−
x
)
d
x
\int_{0}^{4}\left[2^{\frac{x}{2}}+\frac{x}{2}+2\right](4-x) d x
∫
0
4
[
2
2
x
+
2
x
+
2
]
(
4
−
x
)
d
x
\newline
(C)
∫
0
4
[
2
x
2
−
x
2
−
2
]
(
4
−
x
)
d
x
\int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}-2\right](4-x) d x
∫
0
4
[
2
2
x
−
2
x
−
2
]
(
4
−
x
)
d
x
\newline
(D)
∫
0
4
[
2
x
2
−
x
2
+
2
]
(
4
−
x
)
d
x
\int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}+2\right](4-x) d x
∫
0
4
[
2
2
x
−
2
x
+
2
]
(
4
−
x
)
d
x
Get tutor help
The base of a solid is the region enclosed by the graphs of
y
=
x
2
−
5
x
+
7
y=x^{2}-5 x+7
y
=
x
2
−
5
x
+
7
and
y
=
3
y=3
y
=
3
, between
x
=
1
x=1
x
=
1
and
x
=
4
x=4
x
=
4
.
\newline
Cross sections of the solid perpendicular to the
x
x
x
-axis are rectangles whose height is
x
x
x
.
\newline
Which one of the definite integrals gives the volume of the solid?
\newline
Choose
1
1
1
answer:
\newline
(A)
∫
1
4
(
x
2
−
5
x
+
4
)
⋅
x
d
x
\int_{1}^{4}\left(x^{2}-5 x+4\right) \cdot x d x
∫
1
4
(
x
2
−
5
x
+
4
)
⋅
x
d
x
\newline
(B)
π
∫
1
4
(
x
2
−
5
x
+
4
)
2
d
x
\pi \int_{1}^{4}\left(x^{2}-5 x+4\right)^{2} d x
π
∫
1
4
(
x
2
−
5
x
+
4
)
2
d
x
\newline
(C)
∫
1
4
(
x
2
−
5
x
+
4
)
2
d
x
\int_{1}^{4}\left(x^{2}-5 x+4\right)^{2} d x
∫
1
4
(
x
2
−
5
x
+
4
)
2
d
x
\newline
(D)
∫
1
4
(
−
x
2
+
5
x
−
4
)
⋅
x
d
x
\int_{1}^{4}\left(-x^{2}+5 x-4\right) \cdot x d x
∫
1
4
(
−
x
2
+
5
x
−
4
)
⋅
x
d
x
Get tutor help
The base of a solid is the region enclosed by the graphs of
y
=
x
2
−
5
x
+
7
y=x^{2}-5 x+7
y
=
x
2
−
5
x
+
7
and
y
=
3
y=3
y
=
3
, between
x
=
1
x=1
x
=
1
and
x
=
4
x=4
x
=
4
.
\newline
Cross sections of the solid perpendicular to the
x
x
x
-axis are rectangles whose height is
x
x
x
.
\newline
Which one of the definite integrals gives the volume of the solid?
\newline
Choose
1
1
1
answer:
\newline
(A)
∫
1
4
(
x
2
−
5
x
+
4
)
⋅
x
d
x
\int_{1}^{4}\left(x^{2}-5 x+4\right) \cdot x d x
∫
1
4
(
x
2
−
5
x
+
4
)
⋅
x
d
x
\newline
(B)
∫
1
4
(
−
x
2
+
5
x
−
4
)
⋅
x
d
x
\int_{1}^{4}\left(-x^{2}+5 x-4\right) \cdot x d x
∫
1
4
(
−
x
2
+
5
x
−
4
)
⋅
x
d
x
\newline
(C)
∫
1
4
(
x
2
−
5
x
+
4
)
2
d
x
\int_{1}^{4}\left(x^{2}-5 x+4\right)^{2} d x
∫
1
4
(
x
2
−
5
x
+
4
)
2
d
x
\newline
(D)
π
∫
1
4
(
x
2
−
5
x
+
4
)
2
d
x
\pi \int_{1}^{4}\left(x^{2}-5 x+4\right)^{2} d x
π
∫
1
4
(
x
2
−
5
x
+
4
)
2
d
x
Get tutor help
What is the average value of
e
x
e^{x}
e
x
on the interval
[
3
,
9
]
[3,9]
[
3
,
9
]
?
\newline
Choose
1
1
1
answer:
\newline
(A)
e
9
+
e
3
6
\frac{e^{9}+e^{3}}{6}
6
e
9
+
e
3
\newline
(B)
e
9
+
e
3
2
\frac{e^{9}+e^{3}}{2}
2
e
9
+
e
3
\newline
(C)
e
9
−
e
3
6
\frac{e^{9}-e^{3}}{6}
6
e
9
−
e
3
\newline
(D)
e
9
−
e
3
2
\frac{e^{9}-e^{3}}{2}
2
e
9
−
e
3
Get tutor help
The population of a town grows at a rate of
r
(
t
)
r(t)
r
(
t
)
people per year (where
t
t
t
is time in years). At
t
=
3
t=3
t
=
3
, the town's population was
1000
1000
1000
people.
\newline
What does
1000
+
∫
3
8
r
(
t
)
d
t
=
1500
1000+\int_{3}^{8} r(t) d t=1500
1000
+
∫
3
8
r
(
t
)
d
t
=
1500
mean?
\newline
Choose
1
1
1
answer:
\newline
(A) The town's population grew at a rate of
1500
1500
1500
people per year at
t
=
8
t=8
t
=
8
.
\newline
(B) The town's population grew by
1500
1500
1500
people between
t
=
3
t=3
t
=
3
and
t
=
8
t=8
t
=
8
.
\newline
(C) The average rate at which the population grew between
t
=
3
t=3
t
=
3
and
t
=
8
t=8
t
=
8
is
1500
1500
1500
people per year.
\newline
(D) At
t
=
8
t=8
t
=
8
, the town's population was
1500
1500
1500
people.
Get tutor help
Consider the following problem:
\newline
The number of people in a cafeteria is changing at a rate of
r
(
t
)
=
1920
−
160
t
r(t)=1920-160 t
r
(
t
)
=
1920
−
160
t
people per hour (where
t
t
t
is the time in hours). At time
t
=
11.5
t=11.5
t
=
11.5
, there were
60
60
60
people in the cafeteria. How many people were in the cafeteria at hour
12
12
12
.
5
5
5
?
\newline
Which expression can we use to solve the problem?
\newline
Choose
1
1
1
answer:
\newline
(A)
60
+
∫
11.5
12.5
r
(
t
)
d
t
60+\int_{11.5}^{12.5} r(t) d t
60
+
∫
11.5
12.5
r
(
t
)
d
t
\newline
(B)
60
+
∫
11.5
12.5
r
′
(
t
)
d
t
60+\int_{11.5}^{12.5} r^{\prime}(t) d t
60
+
∫
11.5
12.5
r
′
(
t
)
d
t
\newline
(C)
∫
11.5
12.5
r
′
(
t
)
d
t
\int_{11.5}^{12.5} r^{\prime}(t) d t
∫
11.5
12.5
r
′
(
t
)
d
t
\newline
(D)
∫
11.5
12.5
r
(
t
)
d
t
\int_{11.5}^{12.5} r(t) d t
∫
11.5
12.5
r
(
t
)
d
t
Get tutor help
Consider the following problem:
\newline
The depth of the snow on Cam's driveway is increasing at a rate of
r
(
t
)
=
t
+
1
2
r(t)=\frac{t+1}{2}
r
(
t
)
=
2
t
+
1
centimeters per hour (where
t
t
t
is the time in hours). At time
t
=
5
t=5
t
=
5
, the depth of the snow is
9
9
9
centimeters. By how much does the depth of the snow increase between hours
5
5
5
and
8
8
8
?
\newline
Which expression can we use to solve the problem?
\newline
Choose
1
1
1
answer:
\newline
(A)
∫
5
8
r
(
t
)
d
t
\int_{5}^{8} r(t) d t
∫
5
8
r
(
t
)
d
t
\newline
(B)
9
+
∫
5
8
r
(
t
)
d
t
9+\int_{5}^{8} r(t) d t
9
+
∫
5
8
r
(
t
)
d
t
\newline
(C)
r
′
(
8
)
−
9
r^{\prime}(8)-9
r
′
(
8
)
−
9
\newline
(D)
r
′
(
8
)
r^{\prime}(8)
r
′
(
8
)
Get tutor help
Consider the following problem:
\newline
The population of ants in Chloe's ant farm changes at a rate of
r
(
t
)
=
−
15.8
⋅
0.
9
t
r(t)=-15.8 \cdot 0.9^{t}
r
(
t
)
=
−
15.8
⋅
0.
9
t
ants per month (where
t
t
t
is time in months). At time
t
=
0
t=0
t
=
0
, the ant farm's population is
150
150
150
ants. How many ants are in the farm at
t
=
4
t=4
t
=
4
?
\newline
Which expression can we use to solve the problem?
\newline
Choose
1
1
1
answer:
\newline
(A)
∫
3
4
r
(
t
)
d
t
\int_{3}^{4} r(t) d t
∫
3
4
r
(
t
)
d
t
\newline
(B)
150
+
∫
0
4
r
(
t
)
d
t
150+\int_{0}^{4} r(t) d t
150
+
∫
0
4
r
(
t
)
d
t
\newline
(C)
150
+
∫
3
4
r
(
t
)
d
t
150+\int_{3}^{4} r(t) d t
150
+
∫
3
4
r
(
t
)
d
t
\newline
(D)
∫
0
4
r
(
t
)
d
t
\int_{0}^{4} r(t) d t
∫
0
4
r
(
t
)
d
t
Get tutor help
Find
d
2
d
x
2
[
e
7
x
−
4
]
\frac{d^{2}}{d x^{2}}\left[e^{7 x-4}\right]
d
x
2
d
2
[
e
7
x
−
4
]
\newline
Choose
1
1
1
answer:
\newline
(A)
7
e
7
x
−
4
7 e^{7 x-4}
7
e
7
x
−
4
\newline
(B)
49
e
7
x
−
4
49 e^{7 x-4}
49
e
7
x
−
4
\newline
(C)
−
16
e
7
x
-16 e^{7 x}
−
16
e
7
x
\newline
(D)
16
e
7
x
−
4
16 e^{7 x-4}
16
e
7
x
−
4
Get tutor help
Find
d
2
d
x
2
[
e
7
x
−
4
]
\frac{d^{2}}{d x^{2}}\left[e^{7 x-4}\right]
d
x
2
d
2
[
e
7
x
−
4
]
\newline
Choose
1
1
1
answer:
\newline
(A)
49
e
7
x
−
4
49 e^{7 x-4}
49
e
7
x
−
4
\newline
(B)
7
e
7
x
−
4
7 e^{7 x-4}
7
e
7
x
−
4
\newline
(C)
−
16
e
7
x
-16 e^{7 x}
−
16
e
7
x
\newline
(D)
16
e
7
x
−
4
16 e^{7 x-4}
16
e
7
x
−
4
Get tutor help
g
(
x
)
=
sin
(
x
)
g
′
(
x
)
=
?
\begin{array}{l} g(x)=\sqrt{\sin (x)} \\ g^{\prime}(x)=? \end{array}
g
(
x
)
=
sin
(
x
)
g
′
(
x
)
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
cos
(
x
)
\sqrt{\cos (x)}
cos
(
x
)
\newline
(B)
cos
(
x
)
2
x
\frac{\cos (\sqrt{x})}{2 \sqrt{x}}
2
x
c
o
s
(
x
)
\newline
(C)
cos
(
x
)
2
sin
(
x
)
\frac{\cos (x)}{2 \sqrt{\sin (x)}}
2
s
i
n
(
x
)
c
o
s
(
x
)
\newline
(D)
[
sin
(
x
)
]
−
1
2
2
\frac{[\sin (x)]^{-\frac{1}{2}}}{2}
2
[
s
i
n
(
x
)
]
−
2
1
Get tutor help
Let
g
(
x
)
=
e
tan
(
x
)
g(x)=e^{\tan (x)}
g
(
x
)
=
e
t
a
n
(
x
)
.
\newline
Find
g
′
(
x
)
g^{\prime}(x)
g
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
e
tan
(
x
)
e^{\tan (x)}
e
t
a
n
(
x
)
\newline
(B)
e
tan
(
x
)
sec
2
(
x
)
e^{\tan (x)} \sec ^{2}(x)
e
t
a
n
(
x
)
sec
2
(
x
)
\newline
(C)
tan
(
x
)
e
tan
(
x
)
−
1
\tan (x) e^{\tan (x)-1}
tan
(
x
)
e
t
a
n
(
x
)
−
1
\newline
(D)
e
sec
2
(
x
)
e^{\sec ^{2}(x)}
e
s
e
c
2
(
x
)
Get tutor help
Find
d
d
x
(
[
ln
(
x
)
]
3
)
\frac{d}{d x}\left([\ln (x)]^{3}\right)
d
x
d
(
[
ln
(
x
)
]
3
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
3
[
ln
(
x
)
]
2
3[\ln (x)]^{2}
3
[
ln
(
x
)
]
2
\newline
(B)
(
1
x
)
3
\left(\frac{1}{x}\right)^{3}
(
x
1
)
3
\newline
(c)
3
(
1
x
)
2
3\left(\frac{1}{x}\right)^{2}
3
(
x
1
)
2
\newline
(D)
3
[
ln
(
x
)
]
2
x
\frac{3[\ln (x)]^{2}}{x}
x
3
[
l
n
(
x
)
]
2
Get tutor help
Let
g
(
x
)
=
e
tan
(
x
)
g(x)=e^{\tan (x)}
g
(
x
)
=
e
t
a
n
(
x
)
.
\newline
Find
g
′
(
x
)
g^{\prime}(x)
g
′
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
e
sec
2
(
x
)
e^{\sec ^{2}(x)}
e
s
e
c
2
(
x
)
\newline
(B)
e
tan
(
x
)
e^{\tan (x)}
e
t
a
n
(
x
)
\newline
(C)
e
tan
(
x
)
sec
2
(
x
)
e^{\tan (x)} \sec ^{2}(x)
e
t
a
n
(
x
)
sec
2
(
x
)
\newline
(D)
tan
(
x
)
e
tan
(
x
)
−
1
\tan (x) e^{\tan (x)-1}
tan
(
x
)
e
t
a
n
(
x
)
−
1
Get tutor help
Find
d
d
x
(
[
ln
(
x
)
]
3
)
\frac{d}{d x}\left([\ln (x)]^{3}\right)
d
x
d
(
[
ln
(
x
)
]
3
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
3
[
ln
(
x
)
]
2
3[\ln (x)]^{2}
3
[
ln
(
x
)
]
2
\newline
(B)
3
(
1
x
)
2
3\left(\frac{1}{x}\right)^{2}
3
(
x
1
)
2
\newline
(C)
3
[
ln
(
x
)
]
2
x
\frac{3[\ln (x)]^{2}}{x}
x
3
[
l
n
(
x
)
]
2
\newline
(D)
(
1
x
)
3
\left(\frac{1}{x}\right)^{3}
(
x
1
)
3
Get tutor help
d
d
x
[
sin
7
(
x
)
]
=
?
\frac{d}{d x}\left[\sin ^{7}(x)\right]=?
d
x
d
[
sin
7
(
x
)
]
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
7
sin
6
(
x
)
cos
(
x
)
7 \sin ^{6}(x) \cos (x)
7
sin
6
(
x
)
cos
(
x
)
\newline
(B)
7
cos
6
(
x
)
7 \cos ^{6}(x)
7
cos
6
(
x
)
\newline
(C)
7
x
6
cos
(
x
7
)
7 x^{6} \cos \left(x^{7}\right)
7
x
6
cos
(
x
7
)
\newline
(D)
cos
(
7
x
6
)
\cos \left(7 x^{6}\right)
cos
(
7
x
6
)
Get tutor help
Let
f
(
x
)
=
3
x
f(x)=\frac{3}{x}
f
(
x
)
=
x
3
.
\newline
Select the correct description of the one-sided limits of
f
f
f
at
x
=
0
x=0
x
=
0
.
\newline
Choose
1
1
1
answer:
\newline
(A)
lim
x
→
0
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{+}} f(x)=+\infty
lim
x
→
0
+
f
(
x
)
=
+
∞
and
lim
x
→
0
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{-}} f(x)=+\infty
lim
x
→
0
−
f
(
x
)
=
+
∞
\newline
(B)
lim
x
→
0
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{+}} f(x)=+\infty
lim
x
→
0
+
f
(
x
)
=
+
∞
and
lim
x
→
0
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{-}} f(x)=-\infty
lim
x
→
0
−
f
(
x
)
=
−
∞
\newline
(C)
lim
x
→
0
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{+}} f(x)=-\infty
lim
x
→
0
+
f
(
x
)
=
−
∞
and
lim
x
→
0
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{-}} f(x)=+\infty
lim
x
→
0
−
f
(
x
)
=
+
∞
\newline
(D)
lim
x
→
0
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{+}} f(x)=-\infty
lim
x
→
0
+
f
(
x
)
=
−
∞
and
lim
x
→
0
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{-}} f(x)=-\infty
lim
x
→
0
−
f
(
x
)
=
−
∞
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