Suppose $h(x)=2f(2\tan (x))+2f(6+3\sin (x))$.
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 $h(x)=2f(2\tan (x))+2f(6+3\sin (x))$. 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 $f^{\prime }(2\tan (x))\cdot 2{\sec }^2(x)$. At $x=0$, $\tan (x)=0$ and $\sec (x)=1$, so this simplifies to $2f^{\prime }(0)\cdot 2=8f^{\prime }(0)$.

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

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

Step 6 ：Given that $f^{\prime }(0)=4$ and $f^{\prime }(6)=4$, we can substitute these values into the equation to get $h^{\prime }(0)=8\ast 4+6\ast 4=56$.

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