1. **State the problem:** We are given three differentiable functions $f(x)$, $g(x)$, and $h(x)$ where $f(x) = -10x^2 - 6$, the tangent line to $g(x)$ at $x = -2$ is $y = -10x - 8$, and $h = f \circ g$, meaning $h(x) = f(g(x))$. We need to find $h'(-2)$. 2. **Recall the chain rule:** For composition of functions $h(x) = f(g(x))$, the derivative is given by $$ h'(x) = f'(g(x)) \cdot g'(x). $$ This means we need to find $f'(g(-2))$ and $g'(-2)$. 3. **Find $g'(-2)$ from the tangent line:** The tangent line to $g(x)$ at $x = -2$ is $y = -10x - 8$. The slope of this line is the derivative of $g$ at $-2$, so $$ g'(-2) = -10. $$ 4. **Find $g(-2)$ from the tangent line:** The tangent line passes through the point $( -2, g(-2) )$. Substitute $x = -2$ into the tangent line equation: $$ y = -10(-2) - 8 = 20 - 8 = 12. $$ So, $$ g(-2) = 12. $$ 5. **Find $f'(x)$:** Given $f(x) = -10x^2 - 6$, differentiate: $$ f'(x) = -20x. $$ 6. **Evaluate $f'(g(-2))$:** Substitute $g(-2) = 12$: $$ f'(12) = -20 \times 12 = -240. $$ 7. **Calculate $h'(-2)$ using the chain rule:** $$ h'(-2) = f'(g(-2)) \cdot g'(-2) = (-240) \times (-10) = 2400. $$ **Final answer:** $$ h'(-2) = 2400. $$