Let the functions $f(\alpha,x)$ and $g(\alpha)$ be defined as \[f(\alpha,x)=\dfrac{(\frac{x}{2})^\alpha}{x-1}\qquad\qquad\qquad g(\alpha)=\,\dfrac{d^4f}{dx^4}|_{x=2}\] Then $g(\alpha)$ is a polynomial is $\alpha$.  Find the leading coefficient of $g(\alpha)$.
1. We start with the given function:
   \[
   f(\alpha, x) = \frac{\left(\frac{x}{2}\right)^\alpha}{x-1}
   \]
   and the function \( g(\alpha) \) defined as:
   \[
   g(\alpha) = \frac{d^4 f}{dx^4} \bigg|_{x=2}
   \]

2. To find \( g(\alpha) \), we need to differentiate \( f(\alpha, x) \) four times with respect to \( x \). First, we rewrite \( f(\alpha, x) \) in a more convenient form:
   \[
   (x-1) f(\alpha, x) = \left(\frac{x}{2}\right)^\alpha
   \]

3. Differentiate both sides of the equation four times with respect to \( x \):
   \[
   (x-1) f'(\alpha, x) + f(\alpha, x) = \left(\frac{\alpha}{2}\right) \left(\frac{x}{2}\right)^{\alpha-1}
   \]
   \[
   (x-1) f''(\alpha, x) + 2 f'(\alpha, x) = \left(\frac{\alpha (\alpha-1)}{4}\right) \left(\frac{x}{2}\right)^{\alpha-2}
   \]
   \[
   (x-1) f'''(\alpha, x) + 3 f''(\alpha, x) = \left(\frac{\alpha (\alpha-1) (\alpha-2)}{8}\right) \left(\frac{x}{2}\right)^{\alpha-3}
   \]
   \[
   (x-1) f''''(\alpha, x) + 4 f'''(\alpha, x) = \left(\frac{\alpha (\alpha-1) (\alpha-2) (\alpha-3)}{16}\right) \left(\frac{x}{2}\right)^{\alpha-4}
   \]

4. Evaluate these derivatives at \( x = 2 \):
   \[
   g(\alpha) = f''''(\alpha, 2)
   \]
   Substituting \( x = 2 \) into the fourth derivative equation:
   \[
   (2-1) f''''(\alpha, 2) + 4 f'''(\alpha, 2) = \left(\frac{\alpha (\alpha-1) (\alpha-2) (\alpha-3)}{16}\right) \left(\frac{2}{2}\right)^{\alpha-4}
   \]
   Simplifying, we get:
   \[
   f''''(\alpha, 2) + 4 f'''(\alpha, 2) = \frac{\alpha (\alpha-1) (\alpha-2) (\alpha-3)}{16}
   \]

5. Since \( f'''(\alpha, 2) \) is a cubic polynomial in \( \alpha \), it does not affect the leading term of \( g(\alpha) \). Therefore, the leading term of \( g(\alpha) \) is:
   \[
   g(\alpha) = \frac{\alpha (\alpha-1) (\alpha-2) (\alpha-3)}{16}
   \]

6. The leading coefficient of \( g(\alpha) \) is the coefficient of \( \alpha^4 \) in the polynomial:
   \[
   \frac{1}{16}
   \]

The final answer is \(\boxed{\frac{1}{16}}\)