Question: If $|x|<1$ and $y=1+x+x^{2}+\ldots$ to $\infty$, then find the value of $\frac{d y}{d x}$
Solution:
Since $|x|<1$,
$y=1+x+x^{2}+\ldots$ to $\infty$
$=\frac{1}{1-x}$
$\therefore \frac{d y}{d x}=-\frac{1}{(1-x)^{2}} \cdot-1$
$=\frac{1}{(1-x)^{2}}($ Ans $)$
