Question:

Step 1: Understand the differentiability condition at critical points. For $f(x) = g(x)|(x-1)(x-2)\cdots(x-10)| - 2$ to be derivable (differentiable) for all $x \in \mathbb{R}$, it must be differentiable at the points where the absolute value expression changes sign, namely at $x = 1, 2, 3, \ldots, 10$. Step 2: Determine the necessary condition for differentiability at these points. At each point $x = k$ where $k \in \{1, 2, \ldots, 10\}$, the expression $|(x-1)(x-2)\cdots(x-10)|$ has a corner (non-differentiable point). For $f(x)$ to be differentiable at $x = k$, the factor $g(k)$ must equal zero. This ensures that the product $g(x) \cdot |(x-1)(x-2)\cdots(x-10)|$ remains differentiable at these points. Therefore, we require: $$g(1) = g(2) = g(3) = \cdots = g(10) = 0$$ Step 3: Analyze the polynomial $g(x)$ and its roots. We are given that $g(x) = ax^9 + bx^6 + cx^3 + d$, which is a polynomial of degree at most 9. However, from Step 2, we need $g(x)$ to have at least 10 roots: $x = 1, 2, 3, \ldots, 10$. Step 4: Apply the fundamental theorem of algebra. A polynomial of degree $n$ can have at most $n$ real roots. Since $g(x)$ has degree at most 9, it can have at most 9 roots. But we require it to have 10 roots, which is impossible unless $g(x)$ is the zero polynomial. Therefore: $$g(x) \equiv 0 \text{ for all } x \in \mathbb{R}$$ This means $a = b = c = d = 0$. Step 5: Simplify $f(x)$ and find $f'(-1)$. With $g(x) = 0$, we have: $$f(x) = 0 \cdot |(x-1)(x-2)\cdots(x-10)| - 2 = -2$$ Since $f(x) = -2$ is a constant function: $$f'(-1) = 0$$ **Final Answer:** The value of $f'(-1)$ is $\boxed{0}$, which corresponds to **Option 2**.