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

Question

Aicharya · Accepted Answer

Given, $f(x)=\frac{x}{1+x}$ and $ \begin{array}{rlrl} 	ext { and } & g(x) & =f(f(x)) \ 	herefore & g(x) & =f\left(\frac{x}{x+1}ight) \ & =\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} $

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

Solution