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. 
 A: 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.
A: $$f(x)= \frac{(x+1)^2}{(x^2+1)^3}$$
$$f'(x)= \frac{(((x+1)^2)'*(x^2+1)^3)-((x^2+1)^3)'*(x+1)^2}{((x^2+1)^3)^2}$$
Hopefully you can do the rest.
Edit: don't forget the chain rule when differentiating $(x^2+1)^3$
$$g(x)=(x^2+1)^3$$
$$g'(x)=3(x^2+1)^2*(x^2+1)'$$
A: This is not an answer but it is too long for a comment.
The problem being fixed (typo in the book), let me remember you a small trick which makes life much easier when the expression contains products, quotients, powers. It is logarithmic differentiation.
Consider $$y=\frac{(x+1)^2}{(x^2+1)^3}$$ Take logarithms $$\log(y)=2\log(x+1)-3\log(x^2+1)$$ Differentiate $$\frac {y'}y=\frac 2{x+1}-\frac{6x}{x^2+1}=-2\frac{2x^2+3x-1}{(x+1)(x^2+1)}$$ So $$y'=-2\frac{2x^2+3x-1}{(x+1)(x^2+1)}\times \frac{(x+1)^2}{(x^2+1)^3}=-\frac{2(x+1)(2x^2+3x-1)}{(x^2+1)^4}$$
