By the chain rule,
\(g(x)=f(x^2)\implies g'(x)=2xf'(x^2)\)
so
\(g'(2)=2\times2\times f'(2^2)=4f'(4)\)
You were able to find \(h'(2)=f'(f(2))\times f'(2)\), which requires knowing both \(f(2)\) and \(f'(2)\). Do you also happen to know \(f(4)\) and \(f'(4)\)?
