Suppose $h(x)=2 f(2 \tan (x))+2 f(6+3 \sin (x))$. You are also told that $f(0)=40$ and $f(6)=28$ and that $f^{\prime}(0)=4$ and $f^{\prime}(6)=4$. Then $h^{\prime}(0)=$

Step 1 ：Suppose \(h(x)=2 f(2 \tan (x))+2 f(6+3 \sin (x))\). You are also told that \(f(0)=40\) and \(f(6)=28\) and that \(f^{\prime}(0)=4\) and \(f^{\prime}(6)=4\).

Step 2 ：The question asks for the derivative of the function \(h(x)\) at \(x=0\). To find this, we need to use the chain rule of differentiation. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 3 ：First, we will differentiate \(2f(2\tan(x))\) with respect to \(x\). The derivative of \(f(2\tan(x))\) is \(f'(2\tan(x)) \cdot 2\sec^2(x)\). At \(x=0\), \(\tan(x)=0\) and \(\sec(x)=1\), so this simplifies to \(2f'(0) \cdot 2 = 8f'(0)\).

Step 4 ：Next, we will differentiate \(2f(6+3\sin(x))\) with respect to \(x\). The derivative of \(f(6+3\sin(x))\) is \(f'(6+3\sin(x)) \cdot 3\cos(x)\). At \(x=0\), \(\sin(x)=0\) and \(\cos(x)=1\), so this simplifies to \(2f'(6) \cdot 3 = 6f'(6)\).

Step 5 ：Finally, we add these two results together to find \(h'(0) = 8f'(0) + 6f'(6)\).

Step 6 ：Given that \(f'(0) = 4\) and \(f'(6) = 4\), we can substitute these values into the equation to get \(h'(0) = 8*4 + 6*4 = 56\).

Step 7 ：Final Answer: \(h^{\prime}(0)=\boxed{56}\)