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Mathematics
Problem
If
y
=
∑
r
=
1
x
tan
−
1
(
1
1
+
r
+
r
2
)
y = \sum_{r=1}^{x} \tan^{-1} \left( \frac{1}{1+r+r^2} \right)
y
=
∑
r
=
1
x
tan
−
1
(
1
+
r
+
r
2
1
)
, then
d
y
d
x
\frac{dy}{dx}
d
x
d
y
is equal to a.
1
1
+
x
2
\frac{1}{1+x^2}
1
+
x
2
1
b.
1
1
+
(
1
+
x
)
2
\frac{1}{1+(1+x)^2}
1
+
(
1
+
x
)
2
1
c. 0 d. None of these
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