Quotient Rule for first derivative help.

I'm not sure about the result.

f(x)= (x^{3}+x^{2})/(x^{2}-4)

=(x^{2}-4)(3x^{2}+2x) - (x^{3}+x^{2})(2x) / (x^{2}-4)^{2 }

=(3x^{4}+2x^{3}-12x^{2}-8x) - (2x^{4} + 2x^{3}) / (x^{2}-4)^{2 }

=(x^{4}-12x^{2}-8x) / (x^{2}-4)^{2
Would this be the simplified first derivative?}

Re: Quotient Rule for first derivative help.

Re: Quotient Rule for first derivative help.

Thanks. Now for finding a possible critical value for increasing and decreasing intervals, I would take the numerator = 0. I'm having trouble on how I should go about that. I've looked up information on how to do critical values, but each has you factoring it down into (x-1) type terms. Meanwhile I have, x^{4}-12x^{2}-8x = 0. I can't factor out anything more than x.

Re: Quotient Rule for first derivative help.

Let $\displaystyle y= x^2$, so the equation becomes $\displaystyle y^2- 12y- 8= 0$. Solve that for y then solve $\displaystyle x^2= y$ for x.

Re: Quotient Rule for first derivative help.

Quote:

Originally Posted by

**mettler** Thanks. Now for finding a possible critical value for increasing and decreasing intervals, I would take the numerator = 0. I'm having trouble on how I should go about that. I've looked up information on how to do critical values, but each has you factoring it down into (x-1) type terms. Meanwhile I have, x^{4}-12x^{2}-8x = 0. I can't factor out anything more than x.

$\displaystyle x(x^3-12x-8) = 0$

the cubic factor has three real roots, but none are rational ... you will most probably have to use technology to find them.