Quotent rule

[xxjeepdude44xx]

f(x) = (x²+3x+2)/(x²-1)²

f(x) = (x²-1)(2x+3)-(x²+3x+2)(2x)/(x²-1)²

So if I distribute the numerator I get

2x³+3x²-2x-3-2x³+6x²+4x / (x²-1)²

simplify and multiply out the denominator and factor out a negative from the denominator

9x²+2x-3/-x⁴+2x-1

3(x²-1)/-x⁴-1

does this look right?

 

[Wanderer]

Let \(\displaystyle f(x) = \dfrac{x^2+3x+2}{(x^2-1)^2}\)

Then, let \(\displaystyle u = x^2 + 3x + 2\)

So, \(\displaystyle \dfrac{du}{dx} = 2x + 3\)

Let \(\displaystyle v = (x^2 - 1)^2\)

Then, \(\displaystyle \dfrac{dv}{dx} = 2(2x)(x^2 - 1) = 4x(x^2 - 1)\)

Now, dump into the quotient rule formula, and we get:

\(\displaystyle \dfrac{d[\dfrac{u}{v}]}{dx} = \dfrac{v\dfrac{du}{dx} - u{\dfrac{dv}{dx}}}{v^2}\)

\(\displaystyle \dfrac{d[f(x)]}{dx} = \dfrac{[(2x + 3)(x^2 - 1)^2 - (x^2 + 3x + 2)(4x)(x^2 - 1)]}{(x^2 - 1)^4}\)

Can you take it from here?

 

[DrPhil]

f(x) = (x²+3x+2)/(x²-1)².... The denominator is probably NOT squared??

f(x) = [(x²-1)(2x+3)-(x²+3x+2)(2x)]/(x²-1)².... You need more parentheses when typing inline

So if I distribute the numerator I get

[(2x³+3x²-2x-3) - (2x³+6x²+4x)]/ (x²-1)².... All three terms from the second numerator term are negative

simplify and multiply out the denominator and factor out a negative from the denominator

9x²+2x-3/-x⁴+2x-1.... X

3(x²-1)/-x⁴-1.... X

does this look right?

Look again at the numerator when you subtract instead of adding. It will factor nicely. Also factor the denominator using "difference of two squares."

Everything I did assumes the original denominator was not squared - I suspected that because you didn't change it to cubed in your first step.

 

[xxjeepdude44xx]

my apologies.. yes you're correct

(x²+3x+2)/(X²-1)