differentiation of both sides of the eqution

Why did this happen ?

Why did differentiation of x^i become (i + 1)x^(i) instead of (i)x^(i-1)

It was an index shift. $$\sum_{i=0}^\infty ix^{i-1} = \sum_{i=1}^\infty ix^{i-1} = \sum_{j=0}^\infty (j+1)x^j,$$ where in the last sum, we let $j=i-1$.
• @aceminer Yes, you can do that. \begin{align}\sum_{i=0}^\infty ix^{i-1} & = (0)x^{0-1} + \sum_{i=1}^\infty ix^{i-1} \\[0ex] & = 1x^{0}+2x^1+\ldots\\[2ex]\sum_{j=0}^\infty (j+1)x^j & = (0+1)x^0+(1+1)x^1+\ldots \\[0ex] & = 1x^{0}+2x^1+\ldots \end{align} – Graham Kemp Sep 4 '15 at 8:07
By differentiating both sides: $$\frac{d(1+x+x^2+\ldots)}{dx}=1+2x+3x^2+\ldots=\sum_{i=0}^\infty (i+1)x^i$$ The other side $$\frac{d(\frac{1}{(1-x)})}{dx}=\frac{1}{(1-x)^2}$$