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Mathematics
Problem
Suppose
f
f
f
and
g
g
g
are functions having second derivatives
f
′
′
f''
f
′′
and
g
′
′
g''
g
′′
everywhere. If
f
(
x
)
g
(
x
)
=
1
f(x)g(x)=1
f
(
x
)
g
(
x
)
=
1
for all
x
x
x
and
f
′
f'
f
′
and
g
′
g'
g
′
are never zero, then
f
′
′
(
x
)
f
′
(
x
)
−
g
′
′
(
x
)
g
′
(
x
)
\frac{f''(x)}{f'(x)}-\frac{g''(x)}{g'(x)}
f
′
(
x
)
f
′′
(
x
)
−
g
′
(
x
)
g
′′
(
x
)
equals. (a)
−
2
f
′
(
x
)
f
(
x
)
-\frac{2f'(x)}{f(x)}
−
f
(
x
)
2
f
′
(
x
)
(b)
−
2
g
′
(
x
)
g
(
x
)
-\frac{2g'(x)}{g(x)}
−
g
(
x
)
2
g
′
(
x
)
(c)
−
f
′
(
x
)
f
(
x
)
-\frac{f'(x)}{f(x)}
−
f
(
x
)
f
′
(
x
)
(d)
2
f
′
(
x
)
f
(
x
)
\frac{2f'(x)}{f(x)}
f
(
x
)
2
f
′
(
x
)
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