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Let’s check out your problem:
Let
f
(
x
)
=
cos
(
x
)
x
−
2
f(x)=\cos (x) x^{-2}
f
(
x
)
=
cos
(
x
)
x
−
2
.
\newline
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
View step-by-step help
Home
Math Problems
Algebra 1
Write variable expressions for arithmetic sequences
Full solution
Q.
Let
f
(
x
)
=
cos
(
x
)
x
−
2
f(x)=\cos (x) x^{-2}
f
(
x
)
=
cos
(
x
)
x
−
2
.
\newline
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Identify components:
Identify the function components to apply the product rule:
f
(
x
)
=
g
(
x
)
h
(
x
)
f(x) = g(x)h(x)
f
(
x
)
=
g
(
x
)
h
(
x
)
, where
g
(
x
)
=
cos
(
x
)
g(x) = \cos(x)
g
(
x
)
=
cos
(
x
)
and
h
(
x
)
=
x
−
2
h(x) = x^{-2}
h
(
x
)
=
x
−
2
.
Apply product rule:
Apply the product rule:
f
′
(
x
)
=
g
′
(
x
)
h
(
x
)
+
g
(
x
)
h
′
(
x
)
f'(x) = g'(x)h(x) + g(x)h'(x)
f
′
(
x
)
=
g
′
(
x
)
h
(
x
)
+
g
(
x
)
h
′
(
x
)
.
Calculate
g
′
(
x
)
g'(x)
g
′
(
x
)
:
Calculate
g
′
(
x
)
g'(x)
g
′
(
x
)
:
g
′
(
x
)
=
−
sin
(
x
)
g'(x) = -\sin(x)
g
′
(
x
)
=
−
sin
(
x
)
.
Calculate
h
′
(
x
)
h'(x)
h
′
(
x
)
:
Calculate
h
′
(
x
)
h'(x)
h
′
(
x
)
:
h
′
(
x
)
=
−
2
x
−
3
h'(x) = -2x^{-3}
h
′
(
x
)
=
−
2
x
−
3
.
Plug in values:
Plug in
g
′
(
x
)
g'(x)
g
′
(
x
)
and
h
′
(
x
)
h'(x)
h
′
(
x
)
into the product rule:
f
′
(
x
)
=
(
−
sin
(
x
)
)
(
x
−
2
)
+
(
cos
(
x
)
)
(
−
2
x
−
3
)
f'(x) = (-\sin(x))(x^{-2}) + (\cos(x))(-2x^{-3})
f
′
(
x
)
=
(
−
sin
(
x
))
(
x
−
2
)
+
(
cos
(
x
))
(
−
2
x
−
3
)
.
Simplify expression:
Simplify the expression:
f
′
(
x
)
=
−
sin
(
x
)
x
2
−
2
cos
(
x
)
x
3
f'(x) = -\frac{\sin(x)}{x^2} - \frac{2\cos(x)}{x^3}
f
′
(
x
)
=
−
x
2
s
i
n
(
x
)
−
x
3
2
c
o
s
(
x
)
.
Combine terms:
Combine the terms:
f
′
(
x
)
=
−
sin
(
x
)
x
−
2
cos
(
x
)
x
3
f'(x) = \frac{-\sin(x)x - 2\cos(x)}{x^3}
f
′
(
x
)
=
x
3
−
s
i
n
(
x
)
x
−
2
c
o
s
(
x
)
.
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\newline
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\newline
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\newline
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1
1
1
answer:
\newline
(A) It increases.
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(B) It decreases.
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