If f(x)=(3x^2+x)/(3x^2-x) then what is f'(x)?

Answer 1

f(x) = (3x^2+x)/(3x^2-x)
= (3x+1)/(3x-1)

= 1+ 2/(3x-1)

f'(x) = 2d/dx(3x-1)^-1
= 2*3*(3x-1)^-2
= 6/(3x-1)^2

Answer 2

Ok, by the quotient rule
"Bottom times derivative of the top, minus top times the derivative of the bottom, all over the bottom squared." So...
((3x^2-x)(6x + 1) - (3x^2 + x)(6x + 1)) / (3x^2 - x)^2
If you're not required to simplify this, DON'T. If you're taking a test, and simplify when it's not required, you will
1) waste a lot of time that you could better use solving more questions, and possibly not have time to finish
2) increase the chances of making careless algebraic mistakes.
At the risk of 2), I will disregard my own advice, and simplify the above. WARNING: Don't take my word for it. I make mistakes sometimes, anyone does. You'll probably get a few different answers on this one, as the algebra to simplify this is a bit messy, and this online text editor is very awkward. Instead, WORK IT OUT YOURSELF!
That said, here goes ..
(3x^2-x)(6x + 1) = 18x^3 + 3x^2 -6x^2 - x = 18x^3 - 3x^2 - x
(3x^2+x)(6x +1) = 18x^3 + 3x^2 +6x^2+x= 18x^3 + 9x^2 +x
Subtract for your numerator and get .............-12 x^2 - x
Now your denominator is
(3x^2 - x)^2 and I wouldn't expand this because as it is it's a lot easier to set equal to 0 and solve, which is often the next thing you want to do.
Put this together and you get
(-12x^2 - x)/(3x^2 - x)^2
It's tempting at this point to factor out the x out of top and bottom and cancel it off, but I wouldn't do that, because you're usually looking for critical points. So if you cancel off the x's you'd lose a degree of multiplicity.

Answer 3

23

Answer 4

f(x)=9x^4-x^2
=x^2(9x^2-1)
=x^2(3x+1)(3x-1)

Answer 5

f(x)=9x^4-x^2

Answer 6

f(x)=9x^4-x^2