Question 1008975
 If s(x) = {{{1/sqrt(x) + 2sinx}}}, find s''(x).
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s(x) = {{{x^(-1/2) + 2sin(x)}}}
s'(x) = {{{(-0.5)*x^-1.5 + 2cos(x)}}}
s"(x) = {{{(-0.5)*(-1.5)*x^-2.5 - 2sin(x)}}}
s"(x) = {{{.75*x^-2.5 - 2sin(x)}}}