If f(x) = |x - 2| and g(x) = f(f(x)), then for 2 < x < 4, g'(x) equals

-1

We have,

$f(x)=\left\{\begin{aligned} x-2~, & \text { for } 4>x \geq 2 \\ -(x-2), & \text { for } x<2
\end{aligned}\right.$

∴  $g(x)=f(f(x)) $

$\Rightarrow g(x)=\left\{\begin{array}{cl}f(x-2), & \text { for } 2 \leq x<4 \\ f(-(x-2)), & \text { for } x<2\end{array}\right.$

$\Rightarrow g(x)= \begin{cases}-(x-2-2), & \text { for } 2 \leq x<4 \\ -(-x+2-2), & \text { for } 0<x<2\end{cases}$

$\Rightarrow g^{\prime}(x)=\left\{\begin{aligned}-1, & ~~~2<x<4 \\ 1, & ~~~0<x<2\end{aligned}\right.$