Let f be a differentiable function such that f(0) = 0 , f(1) = 3 ,...

Given $g(x) = f(2x+1) \cdot e^{2f(x)}$.

Differentiating $g(x)$ with respect to $x$ using the product rule and chain rule, we get:
$g^{\prime}(x) = \left[ f^{\prime}(2x+1) \cdot 2 \right] e^{2f(x)} + f(2x+1) \left[ e^{2f(x)} \cdot 2f^{\prime}(x) \right]$

Substitute $x = 0$ into the derivative:
$g^{\prime}(0) = 2 f^{\prime}(1) e^{2f(0)} + 2 f(1) e^{2f(0)} f^{\prime}(0)$

We are given $f(0) = 0$, $f(1) = 3$, $f^{\prime}(0) = 4$, and $g^{\prime}(0) = 40$.
Since $f(0) = 0$, we have $e^{2f(0)} = e^0 = 1$.

Substituting these values into the equation for $g^{\prime}(0)$:
$40 = 2 f^{\prime}(1) (1) + 2 (3) (1) (4)$
$40 = 2 f^{\prime}(1) + 24$
$2 f^{\prime}(1) = 16$
$f^{\prime}(1) = 8$

Answer: 8