Find the second order derivatives of the function
tan⁻¹x
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Answer:
Let:
y
=
arctan
x
so that:
x
=
tan
y
differentiate this last equality with respect to
x
:
1
=
sec
2
y
d
y
d
x
Now using the trigonometric inequality:
sec
2
y
=
1
+
tan
2
y
we have:
1
=
(
1
+
tan
2
y
)
d
y
d
x
1
=
(
1
+
x
2
)
d
y
d
x
that is:
d
y
d
x
=
1
1
+
x
2
Differentiate again using the chain rule:
d
2
y
d
x
2
=
d
d
x
1
1
+
x
2
d
2
y
d
x
2
=
−
1
(
1
+
x
2
)
2
d
d
x
(
1
+
x
2
)
d
2
y
d
x
2
=
−
2
x
(
1
+
x
2
)
2
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