Question

If $f\left(x\right)=x-x^2+x^3-x^4+\cdots$ and $|x|<1$, then $f^{\prime }\left(x\right)=$

• A. $\frac{1}{(1-x{)}^2}$
• B. $-\frac{1}{(1-x{)}^2}$
• C. $\frac{1}{(1+x)}$
• D. $\frac{1}{(1+x{)}^2}$

Answer 1

The correct option is D $\frac{1}{(1+x{)}^2}$

Given series is $f\left(x\right)=x-x^2+x^3-x^4+x^5+......$

Given function forms G.P with common ration $-x$

Sum of infinite terms of G.P. $=\frac{a}{1-r}$

$f\left(x\right)=\frac{x}{1-(-x)}f\left(x\right)=\frac{x}{1+x}$

Differentiating the function using division rule,

$f^{\prime }\left(x\right)=\frac{1(1+x)-x(1+0)}{{(1+x)}^2}{f}^{\prime }\left(x\right)=\frac{1}{{(1+x)}^2}$

So, option D is correct.