Solving Derivative Problems: Find y`, y``

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Homework Help Overview

The discussion revolves around finding the first and second derivatives of the function y = (x-2)/(x^2-x+1) and y = ∛(x^2+x+1). Participants are exploring the challenges associated with expressing these derivatives in a factored form suitable for graphing.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants share their attempts at calculating the derivatives using different methods, including the quotient rule and questioning the need for factoring. There are discussions about the correctness of expressions and the implications of their forms for graphing.

Discussion Status

Several participants have provided feedback on the attempts, with some suggesting that factoring may not be necessary for graphing. There is an ongoing exploration of the correct forms of the derivatives and the implications of their structures, with no clear consensus reached.

Contextual Notes

Some participants express confusion over the factoring of quadratic expressions and the implications for the derivatives. There is mention of the importance of understanding the domain of the derivatives and critical points for graphing purposes.

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Homework Statement


y= (x-2)/(x^2-x+1)

Find y` and y``

Homework Equations


y= [tex]\sqrt[3]{x^2+x+1}[/tex]


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:
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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 derivative.
 
SNOOTCHIEBOOCHEE said:
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 derivative.

yeah i know that & that was not my question.
Well thanks for replying :smile:
 
farmd684 said:
yeah i know that & that was not my question.
Well thanks for replying :smile:

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.
 
mutton said:
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.

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
 
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SNOOTCHIEBOOCHEE said:
Well we know that x^2-x+1 = (x-2)(x+1)
I don't know that! I was under the impression that (x-2)(x+1)= x^2-x -2.

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

But when you do, make sure you pay attention domain of that derivative.
 
farmd684 said:

Homework Statement


y= (x-2)/(x^2-x+1)

Find y` and y``

Homework Equations


y= [tex]\sqrt[3]{x^2+x+1}[/tex]


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
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

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 :)
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.
 
HallsofIvy said:
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.

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 :smile:
 

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