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.

Jordan, The derivative of your function is $\frac{1}{2} + \frac{\cos 2x}{2}$. Now note that $\cos 2x = \cos^2 x -\sin ^2 = \cos^2 x -1 -\cos^2 x =2\cos^2x -1$. Rearranging, you get $$\cos^2 x =\frac{\cos 2x}{2} + \frac{1}{2}.$$  

$$
\begin{align*}
\cos 2x = \cos(x+x) & =\cos x \cos x -\sin x \sin x \\
& = \cos^2x -\sin^2x\\
& = \cos^2x -(1-\cos^2x)\qquad\text{because}~\cos^2x + \sin^2 x =1.\\
& = \cos^2x-1+ \cos^2x\\
& = 2\cos^2x-1 
\end{align*}
$$
So, you have $\cos2x = 2\cos^2x -1$, which is the same as $\cos 2x + 1 = 2\cos^2x$. Divide both sides by 2 to get what you want.
