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Mathematics
Problem
Let
y
=
ln
(
(
1
+
x
2
)
3
/
2
)
y = \ln\bigl((1 + x^2)^{3/2}\bigr)
y
=
ln
(
(
1
+
x
2
)
3/2
)
. a. (i) Show that
d
y
d
x
=
3
x
1
+
x
2
\dfrac{dy}{dx} = \dfrac{3x}{1 + x^2}
d
x
d
y
=
1
+
x
2
3
x
. b. (ii) Prove that
d
2
y
d
x
2
=
3
(
1
−
x
2
)
(
1
+
x
2
)
2
\dfrac{d^2y}{dx^2} = \dfrac{3(1 - x^2)}{(1 + x^2)^2}
d
x
2
d
2
y
=
(
1
+
x
2
)
2
3
(
1
−
x
2
)
. c. (iii) Hence state the intervals on which the curve is concave upward and concave downward.
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