Understanding the Problem

Let's delve into a fascinating problem related to functional equations in calculus. We are given the equation:

For all ( x eq 0 ), we have:

[ f'(x) frac{1}{x} cdot frac{1}{f(x)} ]

Deriving the General Form

First, let's rewrite the equation using a positive real number ( a ) so that it also holds as:

[ f'(ax) frac{1}{x} cdot frac{1}{f(1/x)} ]

Assuming a Specific Form

Let's consider the form ( f(x) c cdot x^n ). This form often simplifies complex functional equations. We substitute this form into the equation to see if it holds.

[ f'(x) c cdot n cdot x^{n-1} ]

Substituting these into the original equation, we get:

[ c cdot n cdot x^{n-1} frac{1}{x} cdot frac{1}{c cdot (1/x)^n} ]

Solving for the Coefficients

Now, let's simplify the right-hand side of the equation:

[ c cdot n cdot x^{n-1} frac{1}{x} cdot frac{x^n}{c} ]

[ c cdot n cdot x^{n-1} frac{x^{n-1}}{c} ]

For this equation to hold for all ( x eq 0 ), the exponents and coefficients must match. Therefore, we can equate the coefficients:

[ c cdot n frac{1}{c} ]

[ c^2 cdot n 1 ]

Since ( n ) is a constant, we can derive that:

[ c^2 frac{1}{n} ]

General Solution

Given that ( c^2 frac{1}{n} ), we can express c as:

[ c pm sqrt{frac{1}{n}} ]

Thus, the function ( f(x) ) can be written as:

[ f(x) pm sqrt{frac{1}{n}} cdot x^n ]

This means we have a family of functions that satisfy the given equation for any constant ( a ) and any real number ( n ).

Conclusion

By exploring the functional equation ( f'(x) frac{1}{x} cdot frac{1}{f(x)} ), we have derived a general form that holds for any positive real number ( a ) and any real number ( n ). The functions in this family are:

[ f(x) pm sqrt{frac{1}{n}} cdot x^n ]

This solution showcases the power of mathematical analysis and the importance of functional equations in understanding the behavior of differentiable functions.