## Calculus 10th Edition

The derivative is $\frac{(5s^2+8)(\sqrt s)}{2s}$
Product Rule $g'(s)=((u(s)(v(s))’=u’(s)v(s)+u(s)v’(s))$ $u(s)=\sqrt s ;u’(s)=\frac{1}{2\sqrt s}$ $v(s)=(s^2+8) ;v’(s)=2s$ $g'(s)= (\frac{1}{2\sqrt s})(s^2+8)+(\sqrt s)(2s)=$ $g'(s)= (\frac{1}{2\sqrt s})(s^2+8)+(\frac{2s}{2\sqrt s)}(2s)=$ $(\frac{1}{2\sqrt s})((s^2+8)+4s^2)=$ $\frac{5s^2+8}{2\sqrt s}=\frac{(5s^2+8)(\sqrt s)}{2s}$