Simple use of the chain rule I have the following problem where I feel I am missing something obvious:
I have the function $f(x)=-\ln(x+x^2)$ with $x>0$ and I wish to find the derivative. By using the chain rule I get
$$
f'(x)=-\frac{2x+1}{x^2+x}
$$
But the solution guide I am looking at gives the derivative
$$
-\left(\frac{1}{x}+\frac{1}{1+x}\right)
$$
Am I missing something here?
 A: 1) The Derivative of f(x):
            a) Definition of f(x) and f'(x):
$$
    f(x)=-\ln (x+x^2)
$$
$$
    f'(x)=\frac {d}{dx} (-\ln (x+x^2))
$$
            b) Chain Rule Definition:
$$
\frac {d}{dx} (h(g(x))=h'(g(x))+g'(x)
$$
            c) Finding f'(x):
$$
    f'(x)=\frac {d}{dx} (-\ln (x+x^2))
$$
$$
h(x)=-\ln(x)
$$
$$
h'(x)=-1/x
$$
$$
g(x)=x+x^2
$$
$$
g'(x)=1+2x
$$
$$
f'(x)=\frac {d}{dx} (-\ln(x+x^2))=\frac{-1}{x+x^2}(1+2x)=-\frac{1+2x}{x+x^2}
$$
2) Partial Fraction Decomposition:
       a) Define new function n(x):
$$
n(x)=\frac{1+2x}{x+x^2}
$$
$$
f'(x)=-n(x)
$$
       b) Partial Fraction:
$$
n(x)=\frac{1+2x}{(x)(1+x)}=\frac{A}{x}+\frac{B}{1+x}=\frac{A(1+x)+B(x)}{(x)(1+x)}
$$
$$
n(x)=\frac{A+Ax+Bx}{(x)(1+x)}=\frac{x(A+B)+1(A)}{(x)(1+x)}
$$
$$
x(A+B)=2x, \:\:  1(A)=1
$$
$$
A+B=2, \:\:  1(A)=1
$$
$$
A=1, \:\:  B=1
$$
$$
 n(x)=\frac{1+2x}{x+x^2}=\frac{A}{x}+\frac{B}{1+x}=\frac{1}{x}+\frac{1}{1+x}
$$
       c) Final Answer:
$$
f'(x)=-n(x)
$$
$$
n(x)=\frac{1}{x}+\frac{1}{1+x}
$$
$$
f'(x)=-\left( \frac{1}{x}+\frac{1}{1+x} \right)
$$
Hope this helps,
Good luck
A: $\displaystyle-\frac{2x+1}{x^2+x}=-\frac{x+(x+1)}{x(x+1)}=-\bigg(\frac{x}{x(x+1)}+\frac{x+1}{x(x+1)}\bigg)=-\bigg(\frac{1}{x}+\frac{1}{x+1}\bigg)$
