Question

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

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

Answer 1

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

$f\left(x\right)=x-x^2+x^3-x^4+\cdots \infty$
This is a G.P. with $a=x,r=-x$
$S_{\infty }=\frac{a}{1-r}$
$\Rightarrow f\left(x\right)=\frac{x}{1+x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(1+x)\frac{d}{dx}\left(x\right)-x\frac{d}{dx}(1+x)}{(1+x{)}^2}$
$=\frac{1+x-x}{(1+x{)}^2}$
$=\frac{1}{(1+x{)}^2}$