On implicit and parametric differentiation

Differentiating the above functional equation with respect to x yields:

$2 \cdot f'(2x+1) = h(2x-5) + 2x \cdot h'(2x-5)$

Now, $h(1), h'(1),$ and $f'(7)$ each occur when $x = 3$ . Substituting the given values for $h(1), h'(1)$ gives:

$2 \cdot f'(7) = h(1) + 6 \cdot h'(1) \Rightarrow f'(7) = \frac{h(1) + 6h'(1)}{2} = \frac{2 + (6)(4)}{2} = \boxed{13}.$