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

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(A) Given the series $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. We know that the Maclaurin series expansion for the exponential fun

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$y$

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(A) Given the series $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. We know that the Maclaurin series expansion for the exponential function is $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. Therefore,$y = e^x$. Now,differentiating both sides with respect to $x$,we get: $\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x$. Since $y = e^x$,we have $\frac{dy}{dx} = y$.

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

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