If y = 1 - x + frac{x^2}{2!} - frac{x^3}{3!} | vedclass

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(D) The given series is the Maclaurin expansion of the exponential function $e^{-x}$.$y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \

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$y$

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(D) The given series is the Maclaurin expansion of the exponential function $e^{-x}$.$y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots = e^{-x}$.Now,differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(e^{-x}) = -e^{-x}$.Next,differentiate $\frac{dy}{dx}$ with respect to $x$ to find the second derivative:$\frac{d^2y}{dx^2} = \frac{d}{dx}(-e^{-x}) = -(-e^{-x}) = e^{-x}$.Since $y = e^{-x}$,we have $\frac{d^2y}{dx^2} = y$.

If $y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots$,then $\frac{d^2y}{dx^2} = $

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