Answer
$r'(x) = (4-x)^{-\frac{3}{2}}$
So,
$r'(0) = \frac{1}{8}$
Work Step by Step
$r(x) = \frac{2}{\sqrt (4-x)}$
$r'(x) = 2(-\frac{1}{2})(4-x)^{-\frac{3}{2}}(-1)$
$r'(x) = (4-x)^{-\frac{3}{2}}$
So,
$r'(0) = \frac{1}{8}$
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