Q63 / 160 · AP EAMCET 2014 · Mathematics

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

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

Solution

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 $$ \begin{aligned} g^{\prime}(x) & =\frac{(2 x+1) 1-x(2)}{(2 x+1)^2} \\ & =\frac{1}{(2 x+1)^2} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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