If fx=x1+x,gx=ffx then g'x=

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

Solution:

$\begin{matrix} g (x) = f [f (x)] \\ = f (\frac{x}{1 + x}) \\ = \frac{x / (1 + x)}{1 + x / (1 + x)} \\ = \frac{x}{1 + 2 x} . \\ g' (x) = \frac{(1 + 2 x) 1 - x (2)}{{(1 + 2 x)}^{2}} \\ = \frac{1}{{(2 x + 1)}^{2}} \end{matrix}$

If $f (x) = \frac{x}{1 + x}, g (x) = f (f (x))$ then $g^{'} (x) =$

Solution:

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