Question

If $f\left(x\right)=\frac{x}{1+x}$ and $g\left(x\right)=f\left(f\right(x\left)\right)$ then $g^{\prime }\left(x\right)=$

• A. $\frac{1}{(2x+3{)}^2}$
• B. $\frac{1}{(x+1{)}^2}$
• C. $\frac{1}{x^2}$
• D. $\frac{1}{(2x+1{)}^2}$

Answer 1

The correct option is D $\frac{1}{(2x+1{)}^2}$

Given, $f\left(x\right)=\frac{x}{1+x}$

$\therefore g\left(x\right)=f\left(f\right(x\left)\right)=\frac{1}{1+\frac{1}{f\left(x\right)}}=\frac{1}{1+\frac{x+1}{x}}=\frac{x}{2x+1}$

$\Rightarrow g^{\prime }\left(x\right)=\frac{(2x+1)\cdot 1-x\left(2\right)}{(2x+1{)}^2}=\frac{1}{(2x+1{)}^2}$