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If $y = 1 + x + {{{x^2}} \over {2\,!}} + {{{x^3}} \over {3\,!}} + ..... + {{{x^n}} \over {n\,!}}$, then ${{dy} \over {dx}} = $
Ay
B$y + {{{x^n}} \over {n!}}$
C$y - {{{x^n}} \over {n!}}$
D$y - 1 - {{{x^n}} \over {n!}}$
Solution
(c) $y = 1 + x + \frac{{{x^2}}}{{2\,!}} + \frac{{{x^3}}}{{3\,!}} + …. + \frac{{{x^n}}}{{n\,!}}$
==> $\left[ {(\log \tan x + 1) + \frac{1}{{\tan x\log \tan x}}} \right]$
==> $\frac{{dy}}{{dx}} + \frac{{{x^n}}}{{n\,!}} = 1 + x + \frac{{{x^2}}}{{2\,!}} + ….. + \frac{{{x^n}}}{{n\,!}}$==> $\frac{{dy}}{{dx}} = y – \frac{{{x^n}}}{{n\,!}}$.
==> $\left[ {(\log \tan x + 1) + \frac{1}{{\tan x\log \tan x}}} \right]$
==> $\frac{{dy}}{{dx}} + \frac{{{x^n}}}{{n\,!}} = 1 + x + \frac{{{x^2}}}{{2\,!}} + ….. + \frac{{{x^n}}}{{n\,!}}$==> $\frac{{dy}}{{dx}} = y – \frac{{{x^n}}}{{n\,!}}$.
Standard 11
Physics
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