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

  • question_answer
    If \[y=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+.....\infty ,\]then \[\frac{dy}{dx}=\] [Karnataka CET 1999]

    A)            y

    B)            \[y-1\]

    C)            \[y+1\]

    D)            None of these

    Correct Answer: A

    Solution :

               \[y=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+......\infty \]Þ\[y={{e}^{x}}\]            Differentiating with respect to x, we get \[\frac{dy}{dx}={{e}^{x}}=y\].


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