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

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Question

(c) $y = 1 + x + \frac{{{x^2}}}{{2\,!}} + \frac{{{x^3}}}{{3\,!}} + .... + \frac{{{x^n}}}{{n\,!}}$
==> $\left[ {(\log 	an x + 1) + \frac{1}{{	an

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) $y = 1 + x + \frac{{{x^2}}}{{2\,!}} + \frac{{{x^3}}}{{3\,!}} + .... + \frac{{{x^n}}}{{n\,!}}$
==> $\left[ {(\log 	an x + 1) + \frac{1}{{	an x\log 	an x}}} ight]$
==> $\frac{{dy}}{{dx}} + \frac{{{x^n}}}{{n\,!}} = 1 + x + \frac{{{x^2}}}{{2\,!}} + ..... + \frac{{{x^n}}}{{n\,!}}$==> $\frac{{dy}}{{dx}} = y - \frac{{{x^n}}}{{n\,!}}$.

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

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