1. **State the problem:** Find the first derivative $f'(x)$ and the second derivative $f''(x)$ of the function $$f(x) = -2 \sin\left(\frac{1}{2}x - 1\right).$$ 2. **Recall the derivative rules:** - The derivative of $\sin(u)$ with respect to $x$ is $\cos(u) \cdot u'$, where $u$ is a function of $x$. - The derivative of a constant times a function is the constant times the derivative of the function. 3. **Find the first derivative $f'(x)$:** $$f'(x) = -2 \cdot \frac{d}{dx} \sin\left(\frac{1}{2}x - 1\right) = -2 \cdot \cos\left(\frac{1}{2}x - 1\right) \cdot \frac{d}{dx} \left(\frac{1}{2}x - 1\right).$$ 4. **Calculate the inner derivative:** $$\frac{d}{dx} \left(\frac{1}{2}x - 1\right) = \frac{1}{2}.$$ 5. **Substitute back:** $$f'(x) = -2 \cdot \cos\left(\frac{1}{2}x - 1\right) \cdot \frac{1}{2} = \cancel{-2 \cdot \frac{1}{2}} \cdot \cos\left(\frac{1}{2}x - 1\right) = -1 \cdot \cos\left(\frac{1}{2}x - 1\right) = -\cos\left(\frac{1}{2}x - 1\right).$$ 6. **Find the second derivative $f''(x)$:** $$f''(x) = \frac{d}{dx} \left(-\cos\left(\frac{1}{2}x - 1\right)\right) = - \cdot \frac{d}{dx} \cos\left(\frac{1}{2}x - 1\right) = - \left(-\sin\left(\frac{1}{2}x - 1\right) \cdot \frac{1}{2}\right).$$ 7. **Simplify:** $$f''(x) = - \left(-\sin\left(\frac{1}{2}x - 1\right) \cdot \frac{1}{2}\right) = \cancel{-1 \cdot -\frac{1}{2}} \sin\left(\frac{1}{2}x - 1\right) = \frac{1}{2} \sin\left(\frac{1}{2}x - 1\right).$$ **Final answers:** $$f'(x) = -\cos\left(\frac{1}{2}x - 1\right), \quad f''(x) = \frac{1}{2} \sin\left(\frac{1}{2}x - 1\right).$$