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

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

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$y$

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(D) The given series is the Maclaurin series expansion for the exponential function $e^{x}$.$y = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots = e^{x}$.Now,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(e^{x})$.Since the derivative of $e^{x}$ is $e^{x}$,we have:$\frac{dy}{dx} = e^{x}$.Substituting the original expression for $y$ back into the result,we get:$\frac{dy}{dx} = y$.

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

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