derivative of f[x]= x^2/(x-1)

[confused_07]

Where did I go wrong:

f[x]= x^2/(x-1)
= x^2 (x-1)^-1

Used the product rule:

F'[x]= [x^2(-x+1)^-2] + [2x(x-1)^-1]
= (3x^3-5x^2+2x) / (-x+1)^2(x-1)
= x(3x^2-5x+2) / (-x+1)^2(x+1)
= x(3x-2) / (-x+1)^2

Is this right? Now I have to find the second derivative as I am trying to sketch the graph by hand, identifying all extrema, inflection points, intercepts, and asymptotes. I attempted the second derivative using what I got and it filled up an entire page.

 

[galactus]

First derivative:

\(\displaystyle \L\\\frac{d}{dx}[\frac{x^{2}}{x-1}]\)

Quotient rule:

\(\displaystyle \L\\\frac{(x-1)(2x)-x^{2}}{(x-1)^{2}}=\frac{x^{2}-2x}{(x-1)^{2}}\)


2nd derivative(quotient rule):

\(\displaystyle \L\\\frac{d}{dx}\left[\frac{x^{2}-2x}{(x-1)^{2}}\right]\)

\(\displaystyle \L\\\frac{(x-1)^{2}(2x-2)-(x^{2}-2x)(2)(x-1)}{(x-1)^{4}}\)

\(\displaystyle \L\\\frac{(2x-2)[\overbrace{(x-1)^{2}-(x^{2}-2x)}^{\text{this equals 1}}]}{(x-1)^{4}}\)

\(\displaystyle \L\\\frac{2}{(x-1)^{3}}\)

 

[confused_07]

Thanks. I was under the assumption that the quotient rule was when you had a polynomial ontop.