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Mathematics
Problem
d
d
x
{
e
x
log
(
1
+
x
2
)
}
=
\frac{d}{dx} \{e^x \log(1 + x^2)\} =
d
x
d
{
e
x
lo
g
(
1
+
x
2
)}
=
(a)
e
x
[
log
(
1
+
x
2
)
+
2
x
1
+
x
2
]
e^x \left[\log(1 + x^2) + \frac{2x}{1 + x^2}\right]
e
x
[
lo
g
(
1
+
x
2
)
+
1
+
x
2
2
x
]
(b)
e
x
[
log
(
1
+
x
2
)
−
2
x
1
+
x
2
]
e^x \left[\log(1 + x^2) - \frac{2x}{1 + x^2}\right]
e
x
[
lo
g
(
1
+
x
2
)
−
1
+
x
2
2
x
]
(c)
e
x
[
log
(
1
+
x
2
)
+
x
1
+
x
2
]
e^x \left[\log(1 + x^2) + \frac{x}{1 + x^2}\right]
e
x
[
lo
g
(
1
+
x
2
)
+
1
+
x
2
x
]
(d)
e
x
[
log
(
1
+
x
2
)
−
x
1
+
x
2
]
e^x \left[\log(1 + x^2) - \frac{x}{1 + x^2}\right]
e
x
[
lo
g
(
1
+
x
2
)
−
1
+
x
2
x
]
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