MCQ
If $y = 1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + .....\infty ,$ then ${{dy} \over {dx}} = $
  • $y$
  • B
    $y - 1$
  • C
    $y + 1$
  • D
    None of these

Answer

Correct option: A.
$y$
a
(a) $y = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ......\infty $==>$y = {e^x}$

Differentiating with respect to  $x$ , we get $\frac{{dy}}{{dx}} = {e^x} = y$.

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