136 3 Real Analysis
This follows by applying the product rule to the formula of the determinant. For our
problem, consider the function
f(x)=∣
∣∣
∣∣
∣∣
∣∣
x+a 1 x ··· x
xx+a 2 ··· x
..
...
.
... ..
.
xx···x+an∣
∣∣
∣∣
∣∣
∣∣
.
Its first derivative is
f′(x)=∣∣
∣∣
∣∣
∣
∣∣
11 ··· 1
xx+a 2 ··· x
..
...
.
... ..
.
xx···x+an∣∣
∣∣
∣∣
∣
∣∣
+
∣∣
∣∣
∣∣
∣
∣∣
x+a 1 x··· x
11 ··· 1
..
...
.
... ..
.
xx···x+an∣∣
∣∣
∣∣
∣
∣∣
+···
+
∣∣
∣∣
∣∣
∣∣
∣
x+a 1 x ···x
xx+a 2 ···x
..
...
.
... ..
.
11 ··· 1
∣
∣
∣∣
∣∣
∣∣
∣
.
Proceeding one step further, we see that the second derivative offconsists of two types
of determinants: some that have a row of 0’s, and others that have two rows of 1’s. In
both cases the determinants are equal to zero, showing thatf′′(x)=0. It follows thatf
itself must be a linear function,
f(x)=f( 0 )+f′( 0 )x.One finds immediately thatf( 0 )=a 1 a 2 ···an. To compute
f′( 0 )=∣∣
∣∣
∣∣
∣∣
∣
11 ··· 1
0 a 2 ··· 0
..
...
.
... ..
.
00 ···an∣∣
∣∣
∣∣
∣∣
∣
+
∣∣
∣∣
∣∣
∣∣
∣
a 10 ··· 0
11 ··· 1
..
...
.
... ..
.
00 ···an∣∣
∣∣
∣∣
∣∣
∣
+···+
∣∣
∣∣
∣∣
∣∣
∣
a 10 ··· 0
0 a 2 ··· 0
..
...
.
... ..
.
11 ··· 1
∣∣
∣∣
∣∣
∣∣
∣
expand each determinant along the row of 1’s. The answer is
f′( 0 )=a 2 a 3 ···an+a 1 a 3 ···an+···+a 1 a 2 ···an− 1 ,whence
f(x)=a 1 a 2 ···an[
1 +
(
1
a 1+
1
a 2+···+
1
an)
x]
.
Substitutingx=1, we obtain the formula from the statement.