How would I differentiate cos(2x)/x^1/2

So for this question you can use either the product rule or the quotient rule and I'll run through them both.First the quotient rule:The quotient rule says that if you have h(x)=f(x)/g(x)Then h'(x) = (f'(x)g(x)-f(x)g'(x))/(g(x))^2So using f(x)=cos(2x) and g(x)=x^1/2then f'(x)=-2sin(2x) and g'(x)=1/2x^-1/2Plugging this into our formula gives ush(x) = (-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xAlways remember to simplify afterwards which gives us(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/xSecond the product rule:What the product rule says is that ifh(x) = f(x)g(x)then h'(x) = f(x)g'(x) + f'(x)g(x)So if we say that h(x) = cos(2x)/x^1/2Then we can say that f(x) = cos(2x) and g(x) = x^-1/2Using the product rule we have:f(x) = cos(2x)      f'(x) = -2sin(2x)g(x) = x^-1/2     g'(x) = 1/2x^-3/2So lastly we know that h(x) = f(x)g'(x) + f'(x)g(x)So using what we've found out we can say that h(x) = (cos(2x))/(2x^3/2)-(2sin(2x))/x^1/2Once again simplifying gives us(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x