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

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Question

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

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$y$

Vedclass · Answer

(A) The given series is the Maclaurin series expansion for the exponential function $e^x$. Thus,$y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + .....\infty = e^x$. Differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(e^x)$. Since the derivative of $e^x$ is $e^x$,we have $\frac{dy}{dx} = e^x$. Substituting the original expression for $y$,we get $\frac{dy}{dx} = y$.

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

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