The Art and Craft of Problem Solving

9.3 DIFFERENTIATION AND INTEGRATION 335
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Example 9.3.6 Logarithmic D ifferentiation. Let f(x) = 1] (x+k). Find f'(I).

Solution: Differentiating a product is not that hard , but a more elegant method is
to convert to a sum first by taking logarithms. We have

log(f(x)) = logx+log(x+ 1) + ... +log(x+n),

and differentiation yields

Thus

f'(x ) 1 1 1

--=-+--+ ... +--.

f(x) x x+ 1 x+n

f'(I)= ( n+l)! ( 1 +!+ ... +_

1

_).

2 n+l

Logarithmic differentiation is not just a tool for computing derivatives. It is part of
a larger idea: developing a bank of useful derivatives of "functions of a function" that
you can recognize to analyze the original function. For example, if a problem contains
or can be made to contain the quantity f' (x)/f(x), then antidifferentiation will yield
the logarithm of f(x), which in turn will shed light on f(x). Here is another example
of this style of reasoning.

Example 9.3.7 (Putnam 19 97) Let f be a twice-differentiable real-valued function
satisfying

f(x) + f"(x) = -xg(x)J'(x), (6)

where g(x) � 0 for all real x. Prove that I f(x) I is bounded, i.e., show that there exists
a constantC such that If(x) 1 � C for all x.

Partial Solution: The differential equation cannot easily be solved for f(x), and
integration likewise doesn't seem to help. However, the left-hand side of (6) is similar
to the derivative of something familiar. Observe that

! (f(x)^2 ) = 2f(x)f' (x).

This suggests that we multiply both sides of (6) by f'(x), getting

f(x)J'(x) + f' (x)f" (x) = -xg(x) (f' (x))^2.

Thus

(7)

Now let x � O. The right-hand side of (7) will be nonpositive, which means that
f(x)^2 + f'(x)^2 must be non-increasing for x � O. Hence

f (x)^2 + J' (x)^2 � f(0)^2 + J' (0)^2