(Continued question) The chain rule is \( \frac{d g(h(x))}{d x}=g^{\prime}(h(x)) h^{\prime}(x) \), so without simplification \( f^{\prime}(x)= \) ? \( \frac{1}{2}(\sqrt{x-1})^{-\frac{1}{2}}\left(3 x^{2}\right) \) \( 3(\sqrt{x})^{2}\left(\frac{1}{2} \frac{1}{\sqrt{x}}\right) \) \( 3(x-1)^{2} \frac{1}{2}(\sqrt{x-1})^{-\frac{1}{2}} \) \( \frac{1}{2}\left(\sqrt{x^{3}-1}\right)^{-\frac{1}{2}}\left(3 x^{2}\right) \)

To find \( f'(x) \) using the chain rule, we need to identify the functions involved. Let's assume \( f(x) = g(h(x)) \) where \( g(u) = \sqrt{u} \) and \( h(x) = x^3 - 1 \). Using the chain rule, we have: \[ f'(x) = g'(h(x)) \cdot h'(x) \] 1. **Calculate \( g'(u) \)**: \[ g(u) = \sqrt{u} = u^{1/2} \implies g'(u) = \frac{1}{2} u^{-\frac{1}{2}} \] 2. **Calculate \( h'(x) \)**: \[ h(x) = x^3 - 1 \implies h'(x) = 3x^2 \] 3. **Substituting back into the chain rule**: \[ f'(x) = g'(h(x)) \cdot h'(x) = g'(x^3 - 1) \cdot 3x^2 \] \[ = \frac{1}{2} (x^3 - 1)^{-\frac{1}{2}} \cdot 3x^2 \] Thus, we have: \[ f'(x) = \frac{3}{2} x^2 (x^3 - 1)^{-\frac{1}{2}} \] Now, let's compare this with the options provided: 1. \( \frac{1}{2}(\sqrt{x-1})^{-\frac{1}{2}}\left(3 x^{2}\right) \) 2. \( 3(\sqrt{x})^{2}\left(\frac{1}{2} \frac{1}{\sqrt{x}}\right) \) 3. \( 3(x-1)^{2} \frac{1}{2}(\sqrt{x-1})^{-\frac{1}{2}} \) 4. \( \frac{1}{2}\left(\sqrt{x^{3}-1}\right)^{-\frac{1}{2}}\left(3 x^{2}\right) \) The correct expression for \( f'(x) \) is: \[ \frac{1}{2}\left(\sqrt{x^{3}-1}\right)^{-\frac{1}{2}}\left(3 x^{2}\right) \] Thus, the answer is: \[ \boxed{\frac{1}{2}\left(\sqrt{x^{3}-1}\right)^{-\frac{1}{2}}\left(3 x^{2}\right)} \]