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

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

Vedclass · Accepted Answer

$\frac{1}{25}$

Vedclass · Answer

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

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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