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

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(C) Step $1$: Find the derivatives of the given functions.$f'(x) = \frac{d}{dx}(x^2 + 1)^{1/2} = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt

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

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(C) Step $1$: Find the derivatives of the given functions.$f'(x) = \frac{d}{dx}(x^2 + 1)^{1/2} = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}$.$g'(x) = \frac{(x^2 + 1)(1) - (x + 1)(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2 - 2x}{(x^2 + 1)^2} = \frac{-x^2 - 2x + 1}{(x^2 + 1)^2}$.$h'(x) = \frac{d}{dx}(2x - 3) = 2$.Step $2$: Evaluate the composition $f'(h'(g'(x)))$.Since $h'(x) = 2$ is a constant function, $h'(g'(x)) = 2$ for any $x$.Therefore, $f'(h'(g'(x))) = f'(2)$.Step $3$: Substitute $x = 2$ into $f'(x)$.$f'(2) = \frac{2}{\sqrt{2^2 + 1}} = \frac{2}{\sqrt{5}}$.

If $f(x) = \sqrt{x^2 + 1}$, $g(x) = \frac{x + 1}{x^2 + 1}$, and $h(x) = 2x - 3$, then find the value of $f'(h'(g'(x)))$.

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