Q82 / 180 · KCET 2012 · Mathematics
If $y=e^{\log _{e}\left[1+x+x^{2}+\ldots\right]}$, then $\frac{d y}{d x}$ is equal to
- A$\frac{1}{(1+x)^{2}}$
- B$\frac{1}{(1-x)^{2}}$
- C$\frac{-1}{(1+x)^{2}}$
- Correct answerD$\frac{-1}{(1-x)^{2}}$
Solution
Given, $y=e^{\log \left[1+x+x^{2}+\ldots\right]}$
$$
\begin{array}{cc}
\because & (1-x)^{-1}=1+x+x^{2}+x^{3}+\ldots \infty \\
\therefore & y=e^{\log (1-x)^{-1}} \\
\Rightarrow & y=(1-x)^{-1}
\end{array}
$$
On differentiating w.r.t. ' $x$ ', we get
$$
\begin{aligned}
\frac{d y}{d x} &=(-1)(1-x)^{-2} \\
&=\frac{-1}{(1-x)^{2}}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.