Answer
$h'\left( w \right) = \sqrt 2 $
Work Step by Step
$$\eqalign{
& h\left( w \right) = \sqrt 2 w - \sqrt 2 \cr
& {\text{Differentiate the function}} \cr
& h'\left( w \right) = \frac{d}{{dw}}\left[ {\sqrt 2 w - \sqrt 2 } \right] \cr
& {\text{Use the sum diffence rules for differentiation }} \cr
& h'\left( w \right) = \frac{d}{{dw}}\left[ {\sqrt 2 w} \right] - \frac{d}{{dw}}\left[ {\sqrt 2 } \right] \cr
& {\text{Use the constant multiple rule}} \cr
& h'\left( w \right) = \sqrt 2 \frac{d}{{dw}}\left[ w \right] - \frac{d}{{dw}}\left[ {\sqrt 2 } \right] \cr
& {\text{Apply the rule: }}\frac{d}{{dw}}\left[ w \right] = 1 \cr
& h'\left( w \right) = \sqrt 2 \left( 1 \right) - 0 \cr
& {\text{Rewrite}} \cr
& h'\left( w \right) = \sqrt 2 \cr} $$
