The functions \(f\), \(g\) and \(h\) are related by \[f'(x) = g(x+1), \qquad g'(x)=h(x-1).\] It follows that \(f''(2x)\) equals

\(h(2x+1)\);

\(2h'(2x)\);

\(h(2x)\);

\(4h(2x)\).

Differentiating the first expression gives \[f''(x) = g'(x+1).\] Then by replacing \(x\) with \(x+1\) in the second expression, we have \[g'(x+1)=h(x)\] and so \[f''(x)=h(x).\] Finally, if we replace \(x\) with \(2x\), we have \[f''(2x)=h(2x)\] and so the answer is (c).

You might want to look at A tricky derivative for more on this problem.