first and second derivative of f(x) = (x + 1)^2 / (x - 1)

[Guest]

Hi, can someone check over my work? Thanks

Function:

. . .f(x) = (x + 1)^2 / (x - 1)

First derivative:

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

. . . . . ..= (x^2 - 2x - 3) / (x - 1)^2

Second derivative:

. . .[(2x - 2)(x^2 - 2x + 1) - (x^2 - 2x - 3)(2x - 2)] / (x - 1)^4

. . .[2x^3 - 2x^2 + 2x - 2x^2 + 4x - 2 - 2x^3 + 2x^2 + 4x^2 - 4x + 6x - 6] / (x - 1)^4

. . .(2x^2 + 8x - 6) / (x - 1)^4

I've typed it out in the following image, too:

 

[galactus]

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

Quotient rule:

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

Now, do it again to find the 2nd derivative.

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

\(\displaystyle \L\\\frac{(x-1)^{2}(2x-2)-(x-3)(x+1)(2)(x-1)}{(x-1)^{4}}\)

Simplify:

\(\displaystyle \L\\\frac{8}{(x-1)^{3}}\)