<s>[INST] 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.
 [/INST] $y = \dfrac{1}{2} x + \dfrac {1}{4} \sin(2x)$
$y' = \dfrac{1}{2} + \dfrac{1}{4}\cos(2x)*2$  
$y' = \dfrac{1}{2} + \dfrac{2}{4}\cos(2x)$
$y' = \dfrac{1}{2} + \dfrac{1}{2}\cos(2x)$
$y' = \dfrac{1}{2} \big(1 + \cos (2x)\big) $
Which is equivalent to $\cos^2x$.
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