1. **State the problem:** We want to find the derivative of the function $$f(t) = \left(1 - t\right)^2$$ raised to the power of $$-1$$, which is $$f(t) = \left((1 - t)^2\right)^{-1}$$. 2. **Rewrite the function:** We can write the function as $$f(t) = \frac{1}{(1 - t)^2}$$. 3. **Use the chain rule:** The derivative of $$f(t) = (g(t))^{-1}$$ is $$f'(t) = -1 \cdot (g(t))^{-2} \cdot g'(t)$$ where $$g(t) = (1 - t)^2$$. 4. **Find $$g'(t)$$:** $$g(t) = (1 - t)^2$$ Using the chain rule again, $$g'(t) = 2(1 - t) \cdot (-1) = -2(1 - t)$$. 5. **Apply the chain rule to find $$f'(t)$$:** $$f'(t) = -1 \cdot (1 - t)^{-4} \cdot (-2(1 - t))$$ 6. **Simplify the expression:** $$f'(t) = 2(1 - t) \cdot (1 - t)^{-4} = 2(1 - t)^{1 - 4} = 2(1 - t)^{-3}$$ 7. **Final answer:** $$\boxed{f'(t) = \frac{2}{(1 - t)^3}}$$