If $f(x) = \sqrt{x^2+1}, g(x) = \dfrac{x+1}{x^2+1}, h(x) = 2x - 3$, then $f'\left(h'(g'(x))\right) =$

Given $f(x) = \sqrt{x^2+1}$, differentiating with respect to $x$ gives:

$f'(x) = \dfrac{1}{2\sqrt{x^2+1}} \cdot 2x = \dfrac{x}{\sqrt{x^2+1}}$

Given $h(x) = 2x - 3$, differentiating with respect to $x$ gives:

$h'(x) = 2$

Since $h'(x)$ is a constant function, its value is $2$ for any input. Therefore, for any $x$ where $g'(x)$ is defined:

$h'(g'(x)) = 2$

We need to find $f'(h'(g'(x)))$, which simplifies to $f'(2)$.

Substituting $x = 2$ into the expression for $f'(x)$:

$f'(2) = \dfrac{2}{\sqrt{2^2+1}} = \dfrac{2}{\sqrt{5}}$