# Derivative problem

1. Dec 2, 2008

### farmd684

1. The problem statement, all variables and given/known data
y= (x-2)/(x^2-x+1)

Find y and y

2. Relevant equations
y= $$\sqrt[3]{x^2+x+1}$$

3. The attempt at a solution
for the 1st problem i get to
y=(-4x^2+7x-4)/(x^2-x+1)^2 by u/v method but i have to make it as
-3(x^2-3x+1)/(x^2-x+1)^3
for the second problem
i get to y= (2x+1)/(3(x^2+x+1)^2/3)

then i m messing it up to convert them in a factored forum to solve for x as i have to sketch graphs of them. There are so many problems like this & each time i m feeling dfficulties while making them in a factored form.IS there any intelligent stretegy to do this sort of problems more easily? I will be very glad to know that.
Thanks :)

Last edited: Dec 3, 2008
2. Dec 3, 2008

### SNOOTCHIEBOOCHEE

Well we know that x^2-x+1 = (x-2)(x+1)

some cancel action will help you by making the problem much simpler.

But when you do, make sure you pay attention domain of that derivitive.

3. Dec 3, 2008

### farmd684

yeah i know that & that was not my question.

4. Dec 3, 2008

### mutton

That is the intelligent strategy you are looking for. Read it again.

For the second, you typed y'' but it is y'. There is no more factoring to do, and you don't need to rewrite it to sketch the graph. For example, critical values exist when y' = 0 or y' doesn't exist, so you can look at the numerator and denominator separately.

5. Dec 3, 2008

### farmd684

sorry the second problem will be y' only & i clearly expressed the answer about the 1st question and i want to know how to get into that form.Hope you can understand

Last edited: Dec 3, 2008
6. Dec 3, 2008

### HallsofIvy

Staff Emeritus
I don't know that! I was under the impression that (x-2)(x+1)= x^2-x -2.

7. Dec 3, 2008

### HallsofIvy

Staff Emeritus
You can't. The original y, a linear term over a quadratic, has a "net" power of -1 and so its derivative will have a "net" power of -2. -3(x^2-3x+1)/(x^2-x+1)^3 has a "net" power of 2-6= -4 and cannot possibly be correct. However, I get y'= (-x^2+ 4x- 3)/(x^2-x+1)^2, not your y'.

For the second problem, yes that is correct.
for the second problem

Neither x^2-x+1 nor x^2+x+1 factors with integer coefficients. You can use either completing the square or the quadratic formula to find where they are 0.

8. Dec 3, 2008

### farmd684

for the first problem i get to this
y=(-x^2+4x-1)/(x^2-x+1) i guess this the right one and the answer i got from the book this exactly follows below
y=(-3(x^2-3x+1))/(x^2-x+1)^3 and i m sure about that.Now my question is how to get to this format.
And
for the second problem as you have seen i have done for y and the same format error happenig for me while getting y. I mean factored form like the 1st one.
Thanks for ur time