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

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(C) Given the expression $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$.Differentiating both sides with respect to $x$:$\frac{

Vedclass · Accepted Answer

$y - \frac{x^n}{n!}$

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(C) Given the expression $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$.Differentiating both sides with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}(\frac{x^2}{2!}) + \dots + \frac{d}{dx}(\frac{x^n}{n!})$Since $\frac{d}{dx}(x^k) = k \cdot x^{k-1}$,we have:$\frac{dy}{dx} = 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \dots + \frac{nx^{n-1}}{n!}$Simplifying the terms:$\frac{dy}{dx} = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!}$Now,observe that $y = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{x^n}{n!}$.Therefore,$\frac{dy}{dx} = y - \frac{x^n}{n!}$.

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

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