# Quotient Rule for first derivative help.

• Oct 29th 2012, 04:28 PM
mettler
Quotient Rule for first derivative help.
I'm not sure about the result.

f(x)= (x3+x2)/(x2-4)

=(x2-4)(3x2+2x) - (x3+x2)(2x) / (x2-4)2

=(3x4+2x3-12x2-8x) - (2x4 + 2x3) / (x2-4)2

=(x4-12x2-8x) / (x2-4)2

Would this be the simplified first derivative?
• Oct 29th 2012, 04:34 PM
skeeter
Re: Quotient Rule for first derivative help.
looks ok to me ...
• Oct 29th 2012, 06:38 PM
mettler
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, x4-12x2-8x = 0. I can't factor out anything more than x.
• Oct 30th 2012, 07:52 AM
HallsofIvy
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.
• Oct 30th 2012, 02:26 PM
skeeter
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, x4-12x2-8x = 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.