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

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(C) Given the expression: $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$.To find $\frac{dy}{dx}$,we differentiate each term wi

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

Vedclass · Answer

(C) Given the expression: $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$.To find $\frac{dy}{dx}$,we differentiate each term with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}\left(\frac{x^2}{2!}ight) + \frac{d}{dx}\left(\frac{x^3}{3!}ight) + \dots + \frac{d}{dx}\left(\frac{x^n}{n!}ight)$.Using the power rule $\frac{d}{dx}(x^k) = kx^{k-1}$:$\frac{dy}{dx} = 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \dots + \frac{nx^{n-1}}{n!}$.Since $k! = k 	imes (k-1)!$,we have $\frac{k}{k!} = \frac{1}{(k-1)!}$:$\frac{dy}{dx} = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!}$.Comparing this to the original expression for $y$,we see that the sum is equal to $y$ minus the last term of the original series:$\frac{dy}{dx} = y - \frac{x^n}{n!}$.

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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