Question

If $\left|x\right|<1\mathrm{and}y=1+x+x^2+...$ to ∞, then find the value of $\frac{dy}{dx}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=1+x+x^2+....\mathrm{to}\infty \\ \Rightarrow y=\frac{1}{1-x}\left[\begin{array}{c}\because \mathrm{It}\mathrm{is}\mathrm{a}\mathrm{G}.\mathrm{P}\mathrm{with}\mathrm{first}\mathrm{term}1\\ \mathrm{and}\mathrm{common}\mathrm{ratio}x\end{array}\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{1-x}\right) \\ \Rightarrow \frac{dy}{dx}=-\frac{1}{{\left(1-x\right)}^2}\frac{d}{dx}\left(1-x\right) \\ \Rightarrow \frac{dy}{dx}=-\frac{1}{{\left(1-x\right)}^2}\left(-1\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1}{{\left(1-x\right)}^2} \end{gathered}$$