Question

If $f\left(x\right)=\frac{x}{1+x}$ and $g\left(x\right)=f\left[f\left(x\right)\right]$ then $g\text{'}\left(x\right)$is equal to

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

Answer 1

The correct option is D

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

Explanation for the correct option:

Finding the value of $g\text{'}\left(x\right)$:

Given that,

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

Now,

$\begin{array}{rcl}g\left(x\right)& =& f\left[f\left(x\right)\right]\\ & =& f\left[\frac{x}{1+x}\right]\\ & =& \frac{{\displaystyle \frac{x}{1+x}}}{1+{\displaystyle \frac{x}{1+x}}}\\ & =& \frac{{\displaystyle \frac{x}{1+x}}}{{\displaystyle \frac{2x+1}{1+x}}}\\ & =& \frac{x}{2x+1}\end{array}$

Now, differentiate $g\left(x\right)$ with respect to $x$,

$\begin{array}{rcl}g\text{'}\left(x\right)& =& \frac{\left(2x+1\right)\left(1\right)-x\left(2\right)}{{\left(2x+1\right)}^2}\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu\text{'}-uv\text{'}}{v^2}\right]\\ & =& \frac{2x+1-2x}{{\left(2x+1\right)}^2}\\ & =& \frac{1}{{\left(2x+1\right)}^2}\end{array}$

Hence, the correct option is D.