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Mathematics
Problem
d
d
x
tan
−
1
(
a
x
−
b
b
x
+
a
)
=
\frac{d}{dx} \tan^{-1}\left(\frac{ax - b}{bx + a}\right) =
d
x
d
tan
−
1
(
b
x
+
a
a
x
−
b
)
=
(a)
1
1
+
x
2
−
a
2
a
2
+
b
2
\frac{1}{1 + x^2} - \frac{a^2}{a^2 + b^2}
1
+
x
2
1
−
a
2
+
b
2
a
2
(b)
−
1
1
+
x
2
−
a
2
a
2
+
b
2
\frac{-1}{1 + x^2} - \frac{a^2}{a^2 + b^2}
1
+
x
2
−
1
−
a
2
+
b
2
a
2
(c)
1
1
+
x
2
+
a
2
a
2
+
b
2
\frac{1}{1 + x^2} + \frac{a^2}{a^2 + b^2}
1
+
x
2
1
+
a
2
+
b
2
a
2
(d) None of these
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