## Calculus (3rd Edition)

$h'(4) = \frac{871}{64}$
First, we want to find the derivative using the product rule. Let's let: $u = (s^{-1/2} + 2s)\ and\ v=(7-s^{-1})$ We find the derivatives of each individual part using power rule: $u' = -\frac{1}{2}s^{-3/2} + 2$ $v' = -(-s^{-2}) = s^{-2}$ Now we substitute these values into the equation for product rule: $\frac{d}{dx}(uv) = u'v + v'u$ $h'(s) = (-\frac{1}{2}s^{-3/2} + 2)(7-s^{-1}) +s^{-2}(s^{-1/2} + 2s)$ Now, we need to plug in $s = 4$ to find the derivative at that point: $h'(4) = (-\frac{1}{2}(4)^{-3/2} + 2)(7-4^{-1}) +4^{-2}(4^{-1/2} + 2(4))$ And then we simplify the equation: $h'(4) = (-\frac{1}{2}*\frac{1}{8} + 2)(7-\frac{1}{4}) + \frac{1}{16}(\frac{1}{2} + 8)$ $h'(4) = \frac{31}{16}*\frac{27}{4} + \frac{1}{16}* \frac{17}{2}$ $h'(4) = \frac{837}{64} + \frac{17}{32}$ $h'(4) = \frac{871}{64}$