Question:

<b>Chain Rule for Composite Functions</b><br> Let $h(x) = f(f(f(x)))$ and $g(x) = (f(f(x)))^2$.<br> <b>Part 1:</b> $h'(x) = f'(f(f(x)))\cdot f'(f(x))\cdot f'(x)$.<br> At $x=1$: $f(1)=1$, so $f(f(1))=f(1)=1$, $f(f(f(1)))=1$.<br> $h'(1) = f'(1)\cdot f'(1)\cdot f'(1) = 3^3 = 27$.<br> <b>Part 2:</b> $g'(x) = 2f(f(x))\cdot f'(f(x))\cdot f'(x)$.<br> At $x=1$: $g'(1) = 2\cdot f(1)\cdot f'(1)\cdot f'(1) = 2\cdot 1\cdot 3\cdot 3 = 18$.<br> Total $= 27 + 18 = \boxed{45}$.<br> <b>Key concept:</b> Apply chain rule layer by layer; evaluate each composition at $x=1$ using $f(1)=1$.<br> <b>Trap:</b> Writing $(f(x))^2$ instead of $(f(f(x)))^2$ — the inner composition changes the derivative by a factor of $f'(1)$.