Taking the derivative of $y = \dfrac{x}{2}  + \dfrac {1}{4} \sin(2x)$ Again a simple problem that I can't seem to get the derivative of
I have $\frac{x}{2} + \frac{1}{4}\sin(2x)$
I am getting $\frac{x^2}{4} + \frac{4\sin(2x)}{16}$
This is all very wrong, and I do not know why.

You deal with the sum of functions, $f(x) = \frac{x}{2}$ and $g(x)= \frac{1}{4} \sin(2 x)$. So you would use linearity of the derivative:
$$
  \frac{d}{d x} \left( f(x) + g(x) \right) = \frac{d f(x)}{d x} + \frac{d g(x)}{d x}
$$
To evaluate these derivatives, you would use $\frac{d}{d x}\left( c f(x) \right) = c \frac{d f(x)}{d x}$, for a constant $c$. Thus
$$
 \frac{d}{d x} \left( \frac{x}{2} + \frac{1}{4} \sin(2 x) \right) = \frac{1}{2} \frac{d x}{d x} + \frac{1}{4} \frac{d \sin(2 x)}{d x} 
$$ 
To evaluate derivative of the sine function, you would need a chain rule:
$$
    \frac{d}{d x} y(h(x)) = y^\prime(h(x)) h^\prime(x)
$$
where $y(x) = \sin(x)$ and $h(x) = 2x$. Now finish it off using table of derivatives.
