I have to find the derivative of
f(x) = (3 - 2x - x^2) / (x^2 - 1)
and I get all the way to
(2x^2 - 4x + 2) / ((x^2 - 1)^2)
This is when I get stuck. The book's answer says it is
2 / (x + 1)^2
but I'm having trouble figuring out what to do to get that answer.
f(x) = (3 - 2x - x^2) / (x^2 - 1)
and I get all the way to
(2x^2 - 4x + 2) / ((x^2 - 1)^2)
This is when I get stuck. The book's answer says it is
2 / (x + 1)^2
but I'm having trouble figuring out what to do to get that answer.
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Dear Andrew,
(2x^(2)-4x+2)/((x^(2)-1)^(2))
Factor out the GCF of 2 from each term in the polynomial.
(2(x^(2))+2(-2x)+2(1))/((x^(2)-1)^(2))
Factor out the GCF of 2 from 2x^(2)-4x+2.
(2(x^(2)-2x+1))/((x^(2)-1)^(2))
For a polynomial of the form x^(2)+bx+c, find two factors of c (1) that add up to b (-2). In this problem -1*-1=1 and -1-1=-2, so insert -1 as the right hand term of one factor and -1 as the right-hand term of the other factor.
(2(x-1)(x-1))/((x^(2)-1)^(2))
The binomial can be factored using the difference of squares formula, because both terms are perfect squares. The difference of squares formula is a^(2)-b^(2)=(a-b)(a+b).
(2(x-1)(x-1))/((x-1)(x+1))
Reduce the expression by canceling out the common factor of (x-1) from the numerator and denominator.
(2(x-1)(x-1))/((x-1)…
Reduce the expression by canceling out the common factor of (x-1) from the numerator and denominator.
(2(x-1))/(x+1)
(2x^(2)-4x+2)/((x^(2)-1)^(2))
Factor out the GCF of 2 from each term in the polynomial.
(2(x^(2))+2(-2x)+2(1))/((x^(2)-1)^(2))
Factor out the GCF of 2 from 2x^(2)-4x+2.
(2(x^(2)-2x+1))/((x^(2)-1)^(2))
For a polynomial of the form x^(2)+bx+c, find two factors of c (1) that add up to b (-2). In this problem -1*-1=1 and -1-1=-2, so insert -1 as the right hand term of one factor and -1 as the right-hand term of the other factor.
(2(x-1)(x-1))/((x^(2)-1)^(2))
The binomial can be factored using the difference of squares formula, because both terms are perfect squares. The difference of squares formula is a^(2)-b^(2)=(a-b)(a+b).
(2(x-1)(x-1))/((x-1)(x+1))
Reduce the expression by canceling out the common factor of (x-1) from the numerator and denominator.
(2
Reduce the expression by canceling out the common factor of (x-1) from the numerator and denominator.
(2(x-1))/(x+1)
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this is wrong...you left you the square in the denominator. Maybe you were only doing a general equation but your math is wrong
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you left *out*
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Sorry but the best answer sucks. You are trying to get the derivative of (-x^2-2x+3)/(x^2-1) Then use the quotient rule. y'= [ {f'(x)g(x) - g'(x)f(x)}/ (g(x)^2] --> [ -2x^3-2x^2+2x+2 -(2x^3-4x^2-6x)]/g(x)^2 reduce 2(x-1)^2/(x^2-1)^2---> 2(x-1)^2/ (x-1)^2 (x+1)^2 ---> 2/(x=1)^2
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first you pull out the 2 on the top so the top = 2(x^2 - 2x + 1). then this simplifies into 2(x-1)(x-1) you get this by factoring the equation.
now for the bottom, ((x^2-1)^2) = (x^2 - 1)(x^2 - 1). (x^2 - 1) = (x+1)(x-1) by factoring. so now the bottom equals (x+1)(x-1)(x+1)(x-1).
Now you have (2(x-1)(x-1))/((x-1)(x+1)(x-1)(x+1) so cancel everything out and boom! you got your answer
now for the bottom, ((x^2-1)^2) = (x^2 - 1)(x^2 - 1). (x^2 - 1) = (x+1)(x-1) by factoring. so now the bottom equals (x+1)(x-1)(x+1)(x-1).
Now you have (2(x-1)(x-1))/((x-1)(x+1)(x-1)(x+1) so cancel everything out and boom! you got your answer