Derivatives with different rules I'm having trouble with this one problem that just deals with deriving. I can't seem to figure out how they got their answer. Any help would be appreciated! Thanks!
$
\frac{(x+1)^2}{(x^2+1)^3}
$
The answer is:
$
\frac{-2(x+1)(2x^2-3x-1)}{(x^2+1)^4}
$
I seem to get the wrong answer when using the quotient rule. 

Let's use quotient rule:
$$\frac{(x^2+1)^3\frac{d}{dx}(x+1)^2-(x+1)^2\frac{d}{dx}(x^2+1)^3}{{(x^2+1)^3}^2}$$
Do the derivatives out and simplify the denominator:
$$\frac{(x^2+1)^3\cdot 2(x+1)-(x+1)^2\cdot 2x\cdot 3(x^2+1)^2}{(x^2+1)^6}$$
Factor out $2(x+1)(x^2+1)^2$ from the numerator:
$$\frac{2(x+1)(x^2+1)^2((x^2+1)-(x+1)\cdot 3x)}{(x^2+1)^6}$$
Reduce the fraction and simplify the numerator:
$$\frac{2(x+1)(-2x^2-3x+1)}{(x^2+1)^4}$$
Factor out a $-1$ from $-2x^2-3x+1$:
$$\frac{-2(x+1)(2x^2+3x-1)}{(x^2+1)^4}$$
Thus, it seems that your book's answer is wrong and it should be $+3x$, not $-3x$. Wolfram Alpha also agrees with me on this.
