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.
 A: You want the derivative of a sum of functions, so find the derivative of each term in the sum and add.
For the term ${x\over2}$, note that it can be written as ${1\over2} x$. Now use the fact that you can factor constants out of derivatives:
$$\color{maroon}{
{d\over dx }{x\over 2 } } = 
{d\over dx }\Bigr({1\over 2 } x \Bigl)  ={1\over 2 } {d\over dx } x  ={1\over 2}\cdot 1=\color{maroon}{{1\over2}}.
$$
For the other term, you'll need to use the chain rule and the fact that the derivative of $\sin(x)$ is $\cos(x)$:
$$\color{darkgreen}{
{d\over dx }\Bigr({1\over 4} \sin(2x) \Bigl) }
={1\over 4} {d\over dx } \sin(2x)  = {1\over 4}\cos(2x) \cdot  (2x)' 
= {1\over 4}\cos(2x)   \cdot2=\color{darkgreen}{ {1\over 2}\cos(2x)}.
$$
Combining the above results:
$$
{d\over dx}\Bigl(  {x\over2}+{1\over4}\sin(2x)\Bigr)= 
\color{maroon}{{d\over dx}  {x\over2}}+\color{darkgreen}{ {d\over dx}\Bigr( {1\over4}\sin(2x)\Bigr) }
=\color{maroon}{{1\over2}}+\color{darkgreen}{{1\over 2}\cos(2x) }={1\over2}\bigl(1+\cos(2x)\bigr).
$$
A: 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.
A: 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.
A: If your trying to prove integrals you should make your question a little better, because the current question makes it appear that you do not know how to take a derivative. 
A: If $f(x) = \frac{x}{2} + \frac{1}{4}\sin(2x)$, then $f'(x) =$ $\frac{1}{2} + \frac{cos(2x)}{2}$. If you want $\int{f(x)}$, then we have $\frac{x^2}{4} - \frac{cos(2x)}{8}+C$. Here, we differentiate/integrate $\frac{x}{2}$ and  $\frac{1}{4}\sin(2x)$ separately. Do you have an initial condition?
A: $y = \dfrac{1}{2} x + \dfrac {1}{4} \sin(2x)$
$y' = \dfrac{1}{2} + \dfrac{1}{4}\cos(2x)*2$  
$y' = \dfrac{1}{2} + \dfrac{2}{4}\cos(2x)$
$y' = \dfrac{1}{2} + \dfrac{1}{2}\cos(2x)$
$y' = \dfrac{1}{2} \big(1 + \cos (2x)\big) $
Which is equivalent to $\cos^2x$.
