To find the sum of the given infinite series This is a question from the book 'Differential Calculus' by Joseph Edwards.
Prove that if $x$ be less than unity
$$\frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac{4x^3-8x^7}{1-x^4+x^8} \ldots = \frac{1+2x}{1+x+x^2}$$
As the numerators are negative derivative of the denominators, I tried differentiating the sum of the series
-log{1-x+x^2}-log{1-x^2+x^4}-log{1-x^4+x^8}....but was unable to arrive at the required answer.

Start with the equality
$$
\frac{1+x}{1+x^3}\cdot \frac{1+x^2}{1+x^6}\cdot \frac{1+x^4}{1+x^{12} }\cdots=\frac{(1-x)^{-1}}{(1-x^3)^{-1}}\tag 1
$$
To understand the numerator, note that you can prove by induction that
$$
\prod_{i=0}^{n-1}(1+x^{2^i})=\sum_{i=0}^{2^n-1}x^i
$$
Letting $n\to\infty$, you get
$$
\prod_{i=0}^\infty (1+x^{2^i})=\sum_{i=0}^\infty x^i=\frac1{1-x}\tag2
$$ For the denominator, substitute $x^3$ into $(2)$.
Next, note that $(1+x)/(1+x^3)=1/(1-x+x^2)$, so $(1)$ can be rewritten as
$$
\frac1{1-x+x^2}\cdot \frac1{1-x^2+x^4}\cdot \frac1{1-x^4+x^8}\cdots =\frac{1-x^3}{1-x}=1+x+x^2\tag{3}
$$
Finally, take the logs of both sides in $(3)$, then differentiate. 
