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

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(D) Given,$f(x) = \frac{x}{1+x}$.We are given $g(x) = f(f(x))$.Substituting $f(x)$ into itself:$g(x) = f\left(\frac{x}{1+x}\right) = \frac{\frac{x}{1+

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$\frac{1}{(2x+1)^2}$

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(D) Given,$f(x) = \frac{x}{1+x}$.We are given $g(x) = f(f(x))$.Substituting $f(x)$ into itself:$g(x) = f\left(\frac{x}{1+x}\right) = \frac{\frac{x}{1+x}}{1 + \frac{x}{1+x}}$.Multiplying the numerator and denominator by $(1+x)$:$g(x) = \frac{x}{1+x+x} = \frac{x}{2x+1}$.Now,we differentiate $g(x)$ with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:$g^{\prime}(x) = \frac{(1)(2x+1) - (x)(2)}{(2x+1)^2}$.$g^{\prime}(x) = \frac{2x+1 - 2x}{(2x+1)^2} = \frac{1}{(2x+1)^2}$.

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

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