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

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Question

(a) $y = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ......\infty $==>$y = {e^x}$
Differentiating with respect to x, we get $\frac{{dy}}

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Vedclass · Answer

(a) $y = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ......\infty $==>$y = {e^x}$
Differentiating with respect to x, we get $\frac{{dy}}{{dx}} = {e^x} = y$.

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

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