Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $f(x) = x^5 + x^3$ and let $g(x) = f^{-1}(x)$ be the inverse of $f$. If $g''(2) = \dfrac{a}{b}$ where $a$ and $b$ are positive coprime integers, then the value of $a$ is

Given $f(x) = x^5 + x^3$ and $g(x) = f^{-1}(x)$.

By the definition of inverse functions, we have $f(g(x)) = x$.

Differentiating both sides with respect to $x$, we get:

$f'(g(x)) g'(x) = 1 \Rightarrow g'(x) = \dfrac{1}{f'(g(x))}$

Differentiating again with respect to $x$ using the chain rule and quotient rule:

$g''(x) = -\dfrac{f''(g(x)) g'(x)}{(f'(g(x)))^2}$

Substituting $g'(x) = \dfrac{1}{f'(g(x))}$ into the equation:

$g''(x) = -\dfrac{f''(g(x))}{(f'(g(x)))^3}$

To find $g''(2)$, we first determine $g(2)$. Since $f(1) = 1^5 + 1^3 = 2$, it follows that $g(2) = 1$.

Now, we find the first and second derivatives of $f(x)$ at $x = 1$:

$f'(x) = 5x^4 + 3x^2 \Rightarrow f'(1) = 5(1)^4 + 3(1)^2 = 8$

$f''(x) = 20x^3 + 6x \Rightarrow f''(1) = 20(1)^3 + 6(1) = 26$

Substituting these values into the expression for $g''(2)$:

$g''(2) = -\dfrac{f''(1)}{(f'(1))^3} = -\dfrac{26}{8^3} = -\dfrac{26}{512} = -\dfrac{13}{256}$

Considering the magnitude to match the positive integers condition given in the problem, we have $\left|g''(2)\right| = \dfrac{13}{256}$.

Comparing this with $\dfrac{a}{b}$, we get $a = 13$ and $b = 256$. Since $13$ and $256$ are coprime, the value of $a$ is $13$.

Answer: $13$