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If $f(x)=\frac{x}{1+x}$ and $g(x)=f(f(x))$, then $g^{\prime}(x)$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{(2 x+1)^2}$
Given, $f(x)=\frac{x}{1+x}$
and
$$
\begin{array}{rlrl}
\text { and } & g(x) & =f(f(x)) \\
\therefore & g(x) & =f\left(\frac{x}{x+1}\right) \\
& =\frac{\frac{x}{1+x}}{1+\frac{x}{x+1}} \\
\Rightarrow \quad & g(x) & =\frac{x}{2 x+1}
\end{array}
$$
On differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
g^{\prime}(x) & =\frac{(2 x+1) 1-x(2)}{(2 x+1)^2} \\
& =\frac{1}{(2 x+1)^2}
\end{aligned}
$$
and
$$
\begin{array}{rlrl}
\text { and } & g(x) & =f(f(x)) \\
\therefore & g(x) & =f\left(\frac{x}{x+1}\right) \\
& =\frac{\frac{x}{1+x}}{1+\frac{x}{x+1}} \\
\Rightarrow \quad & g(x) & =\frac{x}{2 x+1}
\end{array}
$$
On differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
g^{\prime}(x) & =\frac{(2 x+1) 1-x(2)}{(2 x+1)^2} \\
& =\frac{1}{(2 x+1)^2}
\end{aligned}
$$
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