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JEE Main & Advanced Mathematics Differentiation Question Bank Derivative at a point Standard Differentiation

  • question_answer
    If \[y=1+x+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{3}}}{3\,!}+.....+\frac{{{x}^{n}}}{n\,!}\], then \[\frac{dy}{dx}=\]

    A)            y

    B)            \[y+\frac{{{x}^{n}}}{n!}\]

    C)            \[y-\frac{{{x}^{n}}}{n!}\]

    D)            \[y-1-\frac{{{x}^{n}}}{n!}\]

    Correct Answer: C

    Solution :

               \[y=1+x+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{3}}}{3\,!}+....+\frac{{{x}^{n}}}{n\,!}\]            Þ  \[\left[ (\log \tan x+1)+\frac{1}{\tan x\log \tan x} \right]\]            Þ \[\frac{dy}{dx}+\frac{{{x}^{n}}}{n\,!}=1+x+\frac{{{x}^{2}}}{2\,!}+.....+\frac{{{x}^{n}}}{n\,!}\]Þ \[\frac{dy}{dx}=y-\frac{{{x}^{n}}}{n\,!}\].


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