If y=e^{log _{e}left[1+x+x^{2}+ldotsright]},t | vedclass

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(B) Given,$y=e^{\log _{e} [1+x+x^{2}+\ldots]}$We know that the sum of an infinite geometric series is given by $1+x+x^{2}+\ldots = \frac{1}{1-x}$ for

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$\frac{1}{(1-x)^{2}}$

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(B) Given,$y=e^{\log _{e} [1+x+x^{2}+\ldots]}$We know that the sum of an infinite geometric series is given by $1+x+x^{2}+\ldots = \frac{1}{1-x}$ for $|x| Therefore,$y = e^{\log _{e} (1-x)^{-1}}$.Using the property $e^{\log _{e} f(x)} = f(x)$,we get $y = (1-x)^{-1}$.Now,differentiating with respect to $x$ using the chain rule:$\frac{d y}{d x} = \frac{d}{d x} (1-x)^{-1} = -1(1-x)^{-2} \cdot \frac{d}{d x}(1-x)$$\frac{d y}{d x} = -1(1-x)^{-2} \cdot (-1) = (1-x)^{-2} = \frac{1}{(1-x)^{2}}$.

If $y=e^{\log _{e}\left[1+x+x^{2}+\ldots\right]}$,then $\frac{d y}{d x}$ is equal to

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