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JEE Main & Advanced Mathematics Differentiation Question Bank Mock Test - Limits and Derivatives

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

    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 :

    [c] \[y=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+...+\frac{{{x}^{n}}}{n!}\] Or \[\frac{dy}{dx}=0+1+x+\frac{{{x}^{2}}}{2!}+...+\frac{{{x}^{n-1}}}{(n-1)!}\] Or \[\frac{dy}{dx}+\frac{{{x}^{n}}}{n!}=1+x+\frac{{{x}^{2}}}{2!}+...+\frac{{{x}^{n}}}{n!}\] Or \[\frac{dy}{dx}=y-\frac{{{x}^{n}}}{n!}\]


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