Question

If $y =e^{\log_e[1+x+x^2+.........]}$, then $\frac {dy}{dx}$=

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

Answer 1

Answer: (D)

Given, $y = e^{log \left[1+x+x^2+\ldots\right]}$
$\because \left(1-x\right)^{-1} = 1+x+x^{2}+x^{3}+\ldots\infty$
$\therefore y = e^{log\left(1-x\right)^{-1}}$
$\Rightarrow y=\left(1-x\right)^{-1}$
On differentiating w.r.t. 'x', we get
$\frac{dy}{dx} = \left(-1\right)\left(1-x\right)^{-2}$
$= \frac{-1}{\left(1-x\right)^{2}}$