Question 4 Suppose $h=f \circ g$. Find $h^{\prime}(0)$ given that $f(0)=6, f^{\prime}(6)=-4, g(0)=6, g^{\prime}(0)=3$. \[ h^{\prime}(0)= \]

Step 1 ：The problem is asking for the derivative of the composite function \(h = f \circ g\) at \(x = 0\).

Step 2 ：The chain rule in calculus states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.

Step 3 ：In this case, the outer function is \(f\) and the inner function is \(g\). So, we need to find \(f'(g(0)) \cdot g'(0)\).

Step 4 ：We are given that \(g'(0) = 3\).

Step 5 ：We are also given that \(f'(6) = -4\) and \(g(0) = 6\). So, \(f'(g(0)) = f'(6) = -4\).

Step 6 ：Substituting these values into the chain rule gives us \(h'(0) = f'(g(0)) \cdot g'(0) = -4 \cdot 3 = -12\).

Step 7 ：Final Answer: The derivative of the composite function \(h = f \circ g\) at \(x = 0\) is \(\boxed{-12}\).