f,
$y'=2.\cos\dfrac{x}{2}.\Big(\dfrac{x}{2}\Big)'-3\sin4x.(4x)'$
$=2.\dfrac{1}{2}.\cos\dfrac{x}{2}-3\sin4x.4$
$=\cos\dfrac{x}{2}-12\sin4x$
k,
$y'=-\dfrac{1}{x^2}-\dfrac{2(x^2)'}{x^4}+\dfrac{3}{2}.\dfrac{(x^4)'}{x^8}$
$=\dfrac{-1}{x^2}-\dfrac{2.2x}{x^4}+\dfrac{3}{2}.\dfrac{4x^3}{x^8}$
$=\dfrac{-1}{x^2}-\dfrac{4}{x^3}+\dfrac{6}{x^5}$