question_answer

If |x|<1 and \[f(x)=1+x+{{x}^{2}}+...\infty ,\] then find the value of f?(x).

Answer:

Given, \[f\,(x)=1+x+{{x}^{2}}+\,\,...\,\,\infty \] \[=\frac{1}{1-x}\]                     [\[\because \]sum of infinite GP]             \[\therefore \]\[f'(x)=\frac{d\,[{{(1-x)}^{-\,1}}]}{d\,(1-x)}\cdot \frac{d\,(1-x)}{dx}\] [on differentiating w.r.t. x]                         \[=-\,1\,{{(1-x)}^{-\,\,1\,\,-\,\,1}}\cdot (0-1)\]      \[\left[ \because \frac{d}{dx}({{x}^{-\,1}})=-\,1\,{{(x)}^{-\,\,1\,\,-\,\,1}} \right]\]                         \[=\frac{-\,1}{{{(1-x)}^{2}}}\cdot (-1)\]                         \[=\frac{1}{{{(1-x)}^{2}}}\]