Question

If $y=e^{\log (1+x+x^2+x^3+\cdots )}$, where $\left|x\right|<1$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{-1}{{(1-x)}^2}$
• B. $\frac{1}{{(1-x)}^2}$
• C. $\frac{1}{{(1+x)}^2}$
• D. None of these

Answer 1

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

We have, $y=e^{\log (1+x+x^2+x^3+\cdots )}$
$\Rightarrow y=1+x+x^2+x^3+\cdots$
$\Rightarrow y=\frac{1}{1-x}={(1-x)}^{-1}$ $[\because S_{\infty }=\frac{a}{1-r}]$
$\Rightarrow \frac{dy}{dx}=(-1){(1-x)}^{-2}(-1)\Rightarrow \frac{dy}{dx}=\frac{1}{{(1-x)}^2}$