Find $f''(2x)$ if $f'(x) = g(x + 1)$ and $g'(x) = h(x - 1)$

Question

Find $f''(2x)$ if $f'(x) = g(x + 1)$ and $g'(x) = h(x - 1)$

Hello, I am stuck on the above problem.

Here's my work:

Differentiate once:

$$\frac{d}{dx} (f(2x)) = 2 \cdot f'(2x) \text{ (Chain Rule)}$$ $$= 2 \cdot g(2x + 1)$$

Differentiate twice:

$$\frac{d}{dx}(2 \cdot g(2x + 1))$$ $$=2 \cdot 2 \cdot h(2x + 1 - 1) \text{ (Chain Rule)}$$ $$= 4 \cdot h(2x)$$

The answer is $$h(2x)$$ (I think that chain rule is not applied in the correct solution.)

This problem makes me feel that my concept is weak :(

Can someone explain why I am wrong?

Thanks

Answer 1

$\frac{d^2}{dx^2}(f(2x))=4f''(2x)$, so if $\frac{d^2}{dx^2}(f(2x))=4h(2x)$, then $f''(2x)=h(2x)$. You applied the chain rule perfectly, but be careful because $\frac{d^2}{dx^2}(f(2x))\not=f''(2x)$