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For the following function:

$$f(x)=\frac{2}{2x^2}-\frac{x}{3}+\frac{4}{5}+\frac{x+1}{x}$$

I got the individual derivatives below:

$$\frac{d}{dx}(\frac{2}{2x^2}) = \frac{d}{dx}(\frac{1}{x^2}) = \frac{-2}{x^3}$$ $$\frac{d}{dx}(\frac{x}{3}) = \frac{3}{9} =\frac{1}{3}$$ $$\frac{d}{dx}(\frac{x+1}{x}) = -\frac{1}{x^2}$$

Then I just put them all togheter like:

$$f'(x)=-\frac{2}{x^3}-\frac{1}{3}-\frac{1}{x^2}$$

But I ran it on WolframAlpha and it says the correct derivative for that function would be:

$$f'(x)=\frac{2}{3}-\frac{2}{x^3}-\frac{1}{x}$$

I'm quite sure I got the individual derivatives correctly. Is it not the right way to do it? Doesn't it work if I add each derivative together, like, there is this sum rule right?

Delta
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    You seem to have made an error when inputting the expression in WA. For me, WA gives the same result as you get. http://www.wolframalpha.com/input/?i=d%2Fdx%28%5Cfrac%7B2%7D%7B2x%5E2%7D-%5Cfrac%7Bx%7D%7B3%7D%2B%5Cfrac%7B4%7D%7B5%7D%2B%5Cfrac%7Bx%2B1%7D%7Bx%7D%29 – Daniel R Dec 12 '13 at 08:01
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    By the way, it looks like you used the quotient rule when taking the derivative of $x/3$. While that gives you the right answer, it's more natural to pull the constant out: $\frac d{dx} (\frac x3) = \frac13 \frac d{dx} (x) = \frac13 \cdot1 = \frac13$. – Greg Martin Dec 12 '13 at 08:17

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