How do I differentiate (x^(3)+2x) / (x^(2)-5)?

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Homework Statement


(x^(3)+2x) / (x^(2)-5) d/dx = ?


Homework Equations


d/dx (f/g) = (g d/dx (f) - f d/dx (g)) / g^2


The Attempt at a Solution



d/dx y = { (x^(2)-5) d/dx (x^(3)+2x) - (x^(3)+2x) d/dx (x^(2)-5) } / (x^(2)-5)^2

This is where it gets complicated for me.

{ (x^(2)-5)(2x^(2)+2) - (x^(3)+2x)(2x) } / (x^(2)-5)^2

I am pretty sure this is at least close, if it is wrong please tell me where I went wrong, and if it is right but not simplified can you explain how to simplify it further.

Thank you in advance, I hope that I gave you appropriate information.
 
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That's almost it. Except d/dx x^(3) is 3x^(2), not 2x^(2). About the only thing you can do to simplify is expand the products in the numerator and add them.
 
The only mistake I noticed was where you have d/dx(x^3 + 2x) = 2x^2 + 2. That should be 3x^2 + 2.

\frac{dy}{dx}~=~\frac{(x^2 - 5)(3x^2 + 2) - (x^3 + 2x)(2x)}{(x^2 - 5)^2}

You might get some simplification if you multiply the two products in the numerator and then combine like terms.
 
Yep that was a careless typing error, thank you for validating.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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