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Beginning
1
Differentiation Review
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User
:
Melikamp/b-calc
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From Wikibooks, open books for an open world
<
User:Melikamp
Differentiation Review
[
edit
|
edit source
]
1. Write the product function
f
(
x
)
{\displaystyle f(x)}
and its rate of change function for
g
(
x
)
=
2
e
5
x
{\displaystyle g(x)=2e^{5}x}
and
h
(
x
)
=
2
x
2
−
x
{\displaystyle h(x)=2x^{2}-x}
.
f
(
x
)
=
2
e
5
x
(
2
x
2
−
x
)
,
f
′
(
x
)
=
2
e
5
(
2
x
2
−
x
)
+
2
e
5
x
(
4
x
−
1
)
{\displaystyle f(x)=2e^{5}x(2x^{2}-x),\ f'(x)=2e^{5}(2x^{2}-x)+2e^{5}x(4x-1)}
f
(
x
)
=
2
e
5
x
(
2
x
2
−
x
)
,
f
′
(
x
)
=
2
e
5
(
2
x
2
−
x
)
+
2
e
5
x
(
4
x
−
1
)
{\displaystyle f(x)=2e^{5}x(2x^{2}-x),\ f'(x)=2e^{5}(2x^{2}-x)+2e^{5}x(4x-1)}
2. Write the product function
f
(
x
)
{\displaystyle f(x)}
and its rate of change function for
g
(
x
)
=
0.5
cos
(
4
x
)
{\displaystyle g(x)=0.5\cos(4x)}
and
h
(
x
)
=
2
ln
(
x
)
{\displaystyle h(x)=2\ln(x)}
.
f
(
x
)
=
cos
(
4
x
)
ln
(
x
)
,
f
′
(
x
)
=
−
4
sin
(
4
x
)
ln
(
x
)
+
cos
(
4
x
)
x
{\displaystyle f(x)=\cos(4x)\ln(x),\ f'(x)=-4\sin(4x)\ln(x)+{\frac {\cos(4x)}{x}}}
f
(
x
)
=
cos
(
4
x
)
ln
(
x
)
,
f
′
(
x
)
=
−
4
sin
(
4
x
)
ln
(
x
)
+
cos
(
4
x
)
x
{\displaystyle f(x)=\cos(4x)\ln(x),\ f'(x)=-4\sin(4x)\ln(x)+{\frac {\cos(4x)}{x}}}
3. Find
f
′
(
x
)
{\displaystyle f'(x)}
if
f
(
x
)
=
(
2
x
+
3
)
e
−
x
{\displaystyle f(x)=(2x+3)e^{-x}}
.
2
e
−
x
−
(
2
x
+
3
)
e
−
x
{\displaystyle 2e^{-x}-(2x+3)e^{-x}}
2
e
−
x
−
(
2
x
+
3
)
e
−
x
{\displaystyle 2e^{-x}-(2x+3)e^{-x}}
4. Find
f
′
(
x
)
{\displaystyle f'(x)}
if
f
(
x
)
=
2
x
3
ln
(
x
)
{\displaystyle f(x)=2{\sqrt {x^{3}}}\ln(x)}
.
3
x
ln
(
x
)
+
2
x
{\displaystyle 3{\sqrt {x}}\ln(x)+2{\sqrt {x}}}
3
x
ln
(
x
)
+
2
x
{\displaystyle 3{\sqrt {x}}\ln(x)+2{\sqrt {x}}}
5. Find
f
′
(
x
)
{\displaystyle f'(x)}
if
f
(
x
)
=
(
x
+
2
)
3
(
1
−
x
2
)
{\displaystyle f(x)=(x+2)^{3}(1-x^{2})}
.
3
(
x
+
2
)
2
(
1
−
x
2
)
−
2
x
(
x
+
2
)
3
{\displaystyle 3(x+2)^{2}(1-x^{2})-2x(x+2)^{3}}
3
(
x
+
2
)
2
(
1
−
x
2
)
−
2
x
(
x
+
2
)
3
{\displaystyle 3(x+2)^{2}(1-x^{2})-2x(x+2)^{3}}
6. Find
f
′
(
x
)
{\displaystyle f'(x)}
if
f
(
x
)
=
e
x
x
{\displaystyle \displaystyle f(x)={\frac {e^{x}}{x}}}
.
x
e
x
−
e
x
x
2
{\displaystyle \displaystyle {\frac {xe^{x}-e^{x}}{x^{2}}}}
x
e
x
−
e
x
x
2
{\displaystyle \displaystyle {\frac {xe^{x}-e^{x}}{x^{2}}}}
