3.2 Continuity, Derivatives, and Integrals 135Solution.We prove that the function
f(t)=at+bt+ctis increasing fort≥0. Its first derivative is
f′(t)=atlna+btlnb+ctlnc,for which we can tell only thatf′( 0 )=lnabc=ln 1=0. However, the second
derivative isf′′(t)=atln^2 a+btln^2 b+ctln^2 c, which is clearly positive. We thus
deduce thatf′is increasing, and sof′(t)≥f′( 0 )=0 fort≥0. Therefore,fitself is
increasing fort≥0, and the conclusion follows.
And now an exciting example found in D. Bu ̧sneag, I. Maftei,Themes for Mathematics
Circles and Contests(Scrisul Românesc, Craiova).
Example.Prove that
∣
∣∣
∣∣
∣
∣∣
∣
1 +a 1 1 ··· 1
11 +a 2 ··· 1
..
...
.
... ..
.
11 ··· 1 +an∣
∣∣
∣∣
∣
∣∣
∣
=a 1 a 2 ···an(
1 +
1
a 1+
1
a 2+···+
1
an)
.
Solution.In general, if the entries of a matrix depend in a differentiable manner on a
parameterx,
⎛
⎜
⎜⎜
⎝
a 11 (x) a 12 (x)···a 1 n(x)
a 21 (x) a 22 (x)···a 2 n(x)
..
...
.
... ..
.
an 1 (x) an 2 (x)···ann(x)⎞
⎟
⎟⎟
⎠
,
then the determinant is a differentiable function ofx, and its derivative is equal to
∣
∣∣
∣
∣∣
∣∣
∣
a′ 11 (x) a′ 12 (x)···a′ 1 n(x)
a 21 (x) a 22 (x)···a 2 n(x)
..
...
.
... ..
.
an 1 (x) an 2 (x)···ann(x)∣
∣∣
∣
∣∣
∣∣
∣
+
∣
∣∣
∣
∣∣
∣∣
∣
a 11 (x) a 12 (x)···a 1 n(x)
a 21 ′(x) a′ 22 (x)···a 2 ′n(x)
..
...
.
... ..
.
an 1 (x) an 2 (x)···ann(x)∣
∣∣
∣
∣∣
∣∣
∣
+···
+
∣∣
∣∣
∣∣
∣∣
∣
∣
a 11 (x) a 12 (x)···a 1 n(x)
a 21 (x) a 22 (x)···a 2 n(x)
..
...
.
... ..
.
an′ 1 (x) an′ 2 (x)..
.a′nn(x)