f(x) = sqrt{x^2 + 1}; g(x) = frac{x + 1}{x^2 | vedclass

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(B) Given functions are $f(x) = \sqrt{x^2 + 1}$,$g(x) = \frac{x + 1}{x^2 + 1}$,and $h(x) = 2x - 3$. First,we find the derivatives of each function: $f

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$\frac{2}{\sqrt{5}}$

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(B) Given functions are $f(x) = \sqrt{x^2 + 1}$,$g(x) = \frac{x + 1}{x^2 + 1}$,and $h(x) = 2x - 3$. First,we find the derivatives of each function: $f'(x) = \frac{d}{dx}(\sqrt{x^2 + 1}) = \frac{1}{2\sqrt{x^2 + 1}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 1}}$. $g'(x) = \frac{d}{dx}(\frac{x + 1}{x^2 + 1}) = \frac{(x^2 + 1)(1) - (x + 1)(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2 - 2x}{(x^2 + 1)^2} = \frac{-x^2 - 2x + 1}{(x^2 + 1)^2}$. $h'(x) = \frac{d}{dx}(2x - 3) = 2$. Now,we evaluate the composite expression $f' [h'(g'(x))]$. Since $h'(x) = 2$ is a constant function,$h'(g'(x)) = 2$ for any value of $x$. Therefore,$f' [h'(g'(x))] = f'(2)$. Substituting $x = 2$ into the expression for $f'(x)$: $f'(2) = \frac{2}{\sqrt{2^2 + 1}} = \frac{2}{\sqrt{4 + 1}} = \frac{2}{\sqrt{5}}$. Thus,the correct option is $B$.

$f(x) = \sqrt{x^2 + 1}; g(x) = \frac{x + 1}{x^2 + 1}; h(x) = 2x - 3$. Then the value of $f' [h'(g'(x))] = $

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