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If $y = 1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + .....\infty ,$then ${{dy} \over {dx}} = $
Ay
B$y - 1$
C$y + 1$
DNone of these
Solution
(a) $y = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ……\infty $==>$y = {e^x}$
Differentiating with respect to x, we get $\frac{{dy}}{{dx}} = {e^x} = y$.
Differentiating with respect to x, we get $\frac{{dy}}{{dx}} = {e^x} = y$.
Standard 11
Physics
