If f(x) = frac{1}{1 + frac{1}{x}} and g(x) = | vedclass

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(B) Given $f(x) = \frac{1}{1 + \frac{1}{x}} = \frac{x}{x + 1}$.Now, $g(x) = \frac{1}{1 + \frac{1}{f(x)}} = \frac{1}{1 + \frac{x + 1}{x}} = \frac{1}{\f

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$\frac{1}{25}$

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(B) Given $f(x) = \frac{1}{1 + \frac{1}{x}} = \frac{x}{x + 1}$.Now, $g(x) = \frac{1}{1 + \frac{1}{f(x)}} = \frac{1}{1 + \frac{x + 1}{x}} = \frac{1}{\frac{x + x + 1}{x}} = \frac{x}{2x + 1}$.To find $g^{\prime}(x)$, we use the quotient rule $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:$g^{\prime}(x) = \frac{(1)(2x + 1) - (x)(2)}{(2x + 1)^2} = \frac{2x + 1 - 2x}{(2x + 1)^2} = \frac{1}{(2x + 1)^2}$.Substituting $x = 2$:$g^{\prime}(2) = \frac{1}{(2(2) + 1)^2} = \frac{1}{(5)^2} = \frac{1}{25}$.

If $f(x) = \frac{1}{1 + \frac{1}{x}}$ and $g(x) = \frac{1}{1 + \frac{1}{f(x)}}$, then $g^{\prime}(2)$ is equal to

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