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Mathematics
Mathematics
Problem
If
y
=
(
1
+
1
x
)
x
y = \left(1 + \frac{1}{x}\right)^x
y
=
(
1
+
x
1
)
x
, then
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
(a)
(
1
+
1
x
)
x
[
log
(
1
+
1
x
)
−
1
1
+
x
]
\left(1 + \frac{1}{x}\right)^x \left[ \log\left(1 + \frac{1}{x}\right) - \frac{1}{1 + x} \right]
(
1
+
x
1
)
x
[
lo
g
(
1
+
x
1
)
−
1
+
x
1
]
(b)
(
1
+
1
x
)
x
[
log
(
1
+
1
x
)
]
\left(1 + \frac{1}{x}\right)^x \left[ \log\left(1 + \frac{1}{x}\right) \right]
(
1
+
x
1
)
x
[
lo
g
(
1
+
x
1
)
]
(c)
(
x
+
1
x
)
x
[
log
(
x
−
1
)
−
x
x
+
1
]
\left(x + \frac{1}{x}\right)^x \left[ \log(x - 1) - \frac{x}{x + 1} \right]
(
x
+
x
1
)
x
[
lo
g
(
x
−
1
)
−
x
+
1
x
]
(d)
(
1
+
1
x
)
x
[
log
(
1
+
1
x
)
+
1
1
+
x
]
\left(1 + \frac{1}{x}\right)^x \left[ \log\left(1 + \frac{1}{x}\right) + \frac{1}{1 + x} \right]
(
1
+
x
1
)
x
[
lo
g
(
1
+
x
1
)
+
1
+
x
1
]
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