An academic question for smart people. Find all functions that satisfy this condition.?

Question

Find all differentiable functions with positive domain and value for which a positive "a" can be found such that :

f'(a/x)=x/f(x)

{" f' " is the derivative of "f"}

Thanks

Answer 1

f'(a/x)= x/f(x)
This is a Differential Equation (nonlinear Ordinary DE):
- Chain Rule doesn't seem to help.
- The fact that the RHS contains 1/f(x) seems to complicate Laplace Analysis.

So let's try to get a handle on it with some ballpark guesses:

Try f(x) = x^n; f'(x)=n x^(n-1)
f'(a/x) = n (a/x)^(n-1)
x/f(x) = x / x^n = 1/ x^(n-1)

This has a solution for:
n (a/x)^(n-1) = 1/x^(n-1)
n a^(n-1)=1
a = n^(1/(n-1)) and a is positive

So f(x) = x^n satisfies this (for n != 0)
Clearly f(x) = C x^n satisfies it too.
So all monomials satisfy it.
In fact if n is positive, it doesn't need to be an integer, it can be any non-zero real.

f(x) = C e^kx doesn't work.
There might well be a more general class of function, but I can't find it.

So my best answer is:
all single-power terms f(x) = C x^p with positive real power p

(This is more general than "all monomials with positive power n")

monomials

Answer 2

f(x) = (x^2)/2 satisfies the requirements if a=2.

Answer 3

x is (five,zero) and y is (zero, - one million) for the inverse of a linear position so the position might have intercepts of x being (zero,five) and y being (- one million,zero) now simply clear up utilizing y = mx + b. We realize b is five and all we have got to clear up for is m m = five/--one million = five so we now have y = 5x + five and the inverse is x = 5y + five - 5y = - x + five y = one million/5x - one million inverse position

Answer 4

Sorry, all I can figure out is that it probably deals with logarithms.