# Question #7fdfd

Jun 2, 2016

(1) $0$

#### Explanation:

Rewriting the original function,

$f \left(x\right) = g \frac{x}{2} + g \frac{- x}{2} + 2 h \left(x\right) + 2 h \left(- x\right)$

Differentiating, and using the chain rule on $g \left(- x\right)$ and $h \left(- x\right)$:

$f ' \left(x\right) = \frac{g ' \left(x\right)}{2} - \frac{g ' \left(- x\right)}{2} + 2 h ' \left(x\right) - 2 h ' \left(- x\right)$

So:

$f ' \left(0\right) = \frac{g ' \left(0\right)}{2} - \frac{g ' \left(0\right)}{2} + 2 h ' \left(0\right) - 2 h ' \left(0\right)$

$f ' \left(0\right) = 0$