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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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