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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.

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1  
Try differentiating the two terms separately and add the results together. The derivative of $x^n$ (for $n\neq 0$) is $n x^{n-1}$, which your text book should have... –  copper.hat Apr 21 '12 at 16:25
    
@copper I know that and that is what I did I am working on proving integrals by finding the derivative. –  user138246 Apr 21 '12 at 16:28
    
Think about your question again @Jordan , you will realize that its awfully simple, –  Tomarinator Apr 21 '12 at 16:35
    
@5tom I have thought about it for 20 minutes now, I don't know what I did wrong. –  user138246 Apr 21 '12 at 16:38
    
derivative of $\frac{x}{2}$ is $\frac{1}{2}$ and derivative of $\frac{1}{4}Sin(2x)$ is $\frac{1}{2}Cos(2x)$ –  Tomarinator Apr 21 '12 at 16:42

6 Answers 6

up vote 2 down vote accepted

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.

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I don't quite follow $$\cos^2 x =\frac{\cos 2x}{2} + \frac{1}{2}.$$ –  user138246 Apr 21 '12 at 17:59
    
@Jordan: Do you understand that $\cos 2x = \cos^2 x -\sin^2 x$? –  Kuku Apr 21 '12 at 18:02
    
No. But what I am curious is to where the 2 in the denominator goes. –  user138246 Apr 21 '12 at 18:12
    
@Jordan: I have added to my post. Hopefully, it helps. –  Kuku Apr 21 '12 at 18:34
    
I still don't get how that is equal to $cos^2 x$ it seems like you are missing a step. –  user138246 Apr 21 '12 at 18:57

$y = \dfrac{1}{2} x + \dfrac {1}{4} \sin(2x)$

$y' = \dfrac{1}{2} + \dfrac{1}{4}\cos(2x)*2$

$y' = \dfrac{1}{2} + \dfrac{2}{4}\cos(2x)$

$y' = \dfrac{1}{2} + \dfrac{1}{2}\cos(2x)$

$y' = \dfrac{1}{2} \big(1 + \cos (2x)\big) $

Which is equivalent to $\cos^2x$.

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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?

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It is suppose to equal$cos^2 x$ –  user138246 Apr 21 '12 at 17:00
    
@Jordan: What is supposed to equal $\cos^2x$? –  Kuku Apr 21 '12 at 17:21
    
The derivative. –  user138246 Apr 21 '12 at 17:26
    
That is not the derivative of your function. –  Ali Apr 21 '12 at 17:29

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: $$ {d\over dx}\Bigl( {x\over2}+{1\over4}\sin(2x)\Bigr)= \color{maroon}{{d\over dx} {x\over2}}+\color{darkgreen}{ {d\over dx}\Bigr( {1\over4}\sin(2x)\Bigr) } =\color{maroon}{{1\over2}}+\color{darkgreen}{{1\over 2}\cos(2x) }={1\over2}\bigl(1+\cos(2x)\bigr). $$

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+1 for the colour. –  LePressentiment Aug 13 '13 at 16:39

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.

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This just seems to needlessly complicate things to an incredibly complex and abstract problem. I know how to get the derivative I just don't know what I did wrong. –  user138246 Apr 21 '12 at 16:35
    
@Jordan Then take a moment to show us what you did, and someone will be glad to help you out. My understanding of your question was that you have difficulty taking derivatives, so I broke it down to application of simple rules, like linearity and chain rule. –  Sasha Apr 21 '12 at 16:36
    
I have no idea what the word linearity means but I am very familiar with the chain rule. –  user138246 Apr 21 '12 at 16:39
    
@Jordan Linearity of the derivative means $\left( c_1 f_1(x) + c_2 f_2(x)\right)^\prime = c_1 f_1^\prime(x) + c_2 f_2^\prime(x)$ for arbitrary constants $c_1$ and $c_2$ and differentiable functions $f_1$ and $f_2$. –  Sasha Apr 21 '12 at 16:40
    
@Jordan Sasha gave a good help. Try and read your notes carefully to understand what theorems and properties of the derivative you need to solve the problem. –  Pedro Tamaroff Apr 21 '12 at 17:01

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.

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