Find all differentiable functions $f:(0,\infty) \to \mathbb{R}$ such that
$$f(b)-f(a)=(b-a)f’(\sqrt{ab}) \qquad \text{for all}\qquad a,b>0.$$

[i]Proposed by Orif Ibrogimov, National University of Uzbekistan[/i]

To find all differentiable functions \( f:(0,\infty) \to \mathbb{R} \) such that 
\[ f(b) - f(a) = (b-a) f'(\sqrt{ab}) \quad \text{for all} \quad a, b > 0, \]
we will proceed through a series of lemmas and logical deductions.

1. **Define the Linear Operator \( T \)**:
   Define \( T: C^1(\mathbb{R}^+) \to C((\mathbb{R}^+)^2) \) as a linear mapping such that:
   \[ T[f](x,y) = f(y) - f(x) - (y-x) f'(\sqrt{xy}) \]
   for all \( x, y > 0 \). The problem is equivalent to characterizing the elements in the kernel of \( T \):
   \[ \ker T = \{ f \mid f(y) - f(x) - (y-x) f'(\sqrt{xy}) = 0, \forall x, y > 0 \}. \]

2. **Lemma 1**: \( 1, x \in \ker T \)
   **Proof**: Direct verification shows that for \( f(x) = 1 \) and \( f(x) = x \), the given functional equation holds.

3. **Lemma 2**: If \( f \in \ker T \), then for \( g(x) = f(1/x) \), \( g \in \ker T \).
   **Proof**: Note that \( g'(x) = -\frac{1}{x^2} f'(1/x) \). If \( f \in \ker T \), then for all \( x, y \):
   \[
   g(y) - g(x) = f(1/y) - f(1/x) = \left( \frac{1}{y} - \frac{1}{x} \right) f'\left( \frac{1}{\sqrt{xy}} \right) = (y-x) \left( -\frac{1}{xy} f'\left( \frac{1}{\sqrt{xy}} \right) \right) = (y-x) g'(\sqrt{xy}).
   \]
   Therefore, \( g \in \ker T \).

4. **Lemma 3**: The set generated by the functions \( 1, x, 1/x \), \( S = \text{span}(1, x, 1/x) \), is a subset of \( \ker T \).
   **Proof**: From Lemma 1, \( x \in \ker T \), and from Lemma 2, \( 1/x \in \ker T \). Since \( \ker T \) is a vector subspace, any linear combination of these functions is also in \( \ker T \).

5. **Lemma 4**: If \( f \in \ker T \) and \( f(p) = f(q) = 0 \) for \( p \neq q \), then \( f(x) = f(pq/x) \) for all \( x > 0 \).
   **Proof**: Given \( f(p) - f(q) = (p-q) f'(\sqrt{pq}) = 0 \), we have \( f'(\sqrt{pq}) = 0 \). For any \( x > 0 \), using \( pq/x \):
   \[
   f(x) - f(pq/x) = (x - pq/x) f'(\sqrt{pq}) = 0,
   \]
   hence \( f(x) = f(pq/x) \) for all \( x > 0 \).

6. **Lemma 5**: If \( f \in \ker T \) and \( f(p) = f(q) = f(r) = 0 \) for three different points \( p, q, r > 0 \), then \( f(x) = 0 \) for all \( x > 0 \).
   **Proof**: From Lemma 4, \( f(x) = f(pq/x) \). Using the other point \( r \), we have \( f(pq/x) = f(pr/x) = f(qr/x) \) for all \( x > 0 \). In particular, \( f(pr/x) = f(qr/x) \). Let \( s = \frac{r\sqrt{pq}}{x} \), then:
   \[
   f(\sqrt{\frac{pq}{s}}) = f(\sqrt{\frac{qp}{s}}).
   \]
   Plugging this into the equation, we get:
   \[
   0 = f(\sqrt{\frac{pq}{s}}) - f(\sqrt{\frac{qp}{s}}) = s \left( \sqrt{\frac{pq}{s}} - \sqrt{\frac{qp}{s}} \right) f'(s).
   \]
   Since \( s > 0 \) and \( \sqrt{\frac{pq}{s}} - \sqrt{\frac{qp}{s}} = \frac{p-q}{\sqrt{pq}} \neq 0 \), it follows that \( f'(s) = 0 \), meaning \( f \) is constant and equal to 0.

7. **Lemma 6**: (Interpolation) There is a unique element \( f \in S \) that passes through the points \( (p, p'), (q, q'), (r, r') \) for any distinct numbers \( p, q, r > 0 \).
   **Proof**: Let \( f = a + bx + c/x \). We need to solve the system:
   \[
   \begin{cases}
   f(p) = p' \\
   f(q) = q' \\
   f(r) = r'
   \end{cases}
   \]
   This translates to solving the matrix equation \( M \mathbf{x} = \mathbf{f} \) where:
   \[
   M = \begin{bmatrix}
   1 & p & \frac{1}{p} \\
   1 & q & \frac{1}{q} \\
   1 & r & \frac{1}{r}
   \end{bmatrix}, \quad \mathbf{x} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}, \quad \mathbf{f} = \begin{pmatrix} p' \\ q' \\ r' \end{pmatrix}.
   \]
   We need to show that \( M \) is non-singular. The determinant of \( M \) is:
   \[
   \det(M) = \begin{vmatrix}
   1 & p & \frac{1}{p} \\
   1 & q & \frac{1}{q} \\
   1 & r & \frac{1}{r}
   \end{vmatrix} = (q-p)(r-p)(r-q) \frac{1}{pqr} \neq 0.
   \]
   Therefore, \( M \) is non-singular, and the matrix equation has a unique solution, implying a unique \( f \in S \) that passes through the given points.

8. **Theorem**: \( \ker T = S \)
   **Proof**: (<=) Direct from Lemma 3. (=>) Let \( f \in \ker T \). From Lemma 6, there is a unique function \( g \in S \) such that \( g(p) = f(p), g(q) = f(q), g(r) = f(r) \). Let \( h = f - g \). Then \( h(p) = h(q) = h(r) = 0 \). From Lemma 5, \( h(x) = 0 \) for all \( x > 0 \), hence \( f(x) = g(x) \) for all \( x > 0 \), meaning \( f \in S \).

Therefore, the solution set of functions that satisfy the original functional equation is:
\[ f(x) = a + bx + c/x. \]

The final answer is \( \boxed{ f(x) = a + bx + c/x } \).