Find the derivative of \( f(x)=\frac{x}{\sqrt{x-1}} \), o. None of these options. ob. \( \frac{x-2}{2(x-1)^{\frac{3}{2}}} \). o. \( \frac{1}{2}(x-1)^{-\frac{1}{2}} \). od. \( \frac{x}{2 \sqrt{x-1}} \). e. \( \frac{x-2}{2 \sqrt{x-1}} \).

To find the derivative of the function \( f(x) = \frac{x}{\sqrt{x-1}} \), we can use the quotient rule. The quotient rule states that if you have a function \( f(x) = \frac{g(x)}{h(x)} \), then the derivative \( f'(x) \) is given by: \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \] In our case, let \( g(x) = x \) and \( h(x) = \sqrt{x-1} \). First, we need to find \( g'(x) \) and \( h'(x) \): 1. \( g'(x) = 1 \) 2. To find \( h'(x) \), we use the chain rule: \[ h(x) = (x-1)^{\frac{1}{2}} \implies h'(x) = \frac{1}{2}(x-1)^{-\frac{1}{2}} \cdot 1 = \frac{1}{2\sqrt{x-1}} \] Now we can apply the quotient rule: \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \] Substituting in the values we found: \[ f'(x) = \frac{(1)(\sqrt{x-1}) - (x)\left(\frac{1}{2\sqrt{x-1}}\right)}{(\sqrt{x-1})^2} \] Now simplify the expression: 1. The denominator: \[ (\sqrt{x-1})^2 = x-1 \] 2. The numerator: \[ \sqrt{x-1} - \frac{x}{2\sqrt{x-1}} = \frac{2(x-1) - x}{2\sqrt{x-1}} = \frac{2x - 2 - x}{2\sqrt{x-1}} = \frac{x - 2}{2\sqrt{x-1}} \] Putting it all together, we have: \[ f'(x) = \frac{\frac{x - 2}{2\sqrt{x-1}}}{x-1} = \frac{x - 2}{2\sqrt{x-1}(x-1)} \] This can be rewritten as: \[ f'(x) = \frac{x - 2}{2(x-1)^{\frac{3}{2}}} \] Thus, the derivative of \( f(x) = \frac{x}{\sqrt{x-1}} \) is: \[ \frac{x - 2}{2(x-1)^{\frac{3}{2}}} \] The correct answer is: **ob. \( \frac{x-2}{2(x-1)^{\frac{3}{2}}} \)**.