Simple use of the chain rule

Question

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?

Answer 1

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

Answer 2

$\displaystyle-\frac{2x+1}{x^2+x}=-\frac{x+(x+1)}{x(x+1)}=-\bigg(\frac{x}{x(x+1)}+\frac{x+1}{x(x+1)}\bigg)=-\bigg(\frac{1}{x}+\frac{1}{x+1}\bigg)$