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$ f(x) = -\displaystyle\frac{x^3+3x+2}{x^2 - 1} $

i know i have to use the quotient rule which gives me this :

$ f(x) = \displaystyle\frac{(x^2-1)*(-2x-3)-(x^2-3x-2)*(2x)}{(x-1)^4} $

how do i proceed now? I want to find the Monotones of this

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    I believe you got your derivative of $f$ wrong – 2017-01-19

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after the Quotient rule ´$$\frac{u'v-uv'}{v^2}$$ we get $$f'(x)=-\frac{(3x^2+3)(x^2-1)-(x^3+3x+2)2x}{(x^2-1)^2}$$ simplifying this we get $$f'(x)=-\frac{3x^4+3x^2-3x^2-3-2x^4-6x^2-4x}{(x^2-1)^2}$$

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    can you simplify this? – 2017-01-19
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    i will try it now thank you a lot – 2017-01-19
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    we can compare the solutions – 2017-01-19
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    alright i think i get it now. How do i proceed now if i want to find out the gradient? – 2017-01-19
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You can usefully do a few things to simplify a bit before taking the derivative - for example $x^3+3x+2=x(x^2-1)+4x+2$ so that $$-f(x)=x+\frac {4x+2}{x^2-1}$$Now use partial fractions to obtain $$f(x)=-x-\frac 3{x-1}-\frac 1{x+1}$$ and then take the derivative to obtain $$f'(x)=-1+\frac 3{(x-1)^2}+\frac 1{(x+1)^2}$$

I would find that easier to sketch, for example, and it is easy to find further derivatives if you need them.