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J & K CET Engineering J and K - CET Engineering Solved Paper-2013

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

    A)  \[{{e}^{x}}\]

    B)  \[\sin \,x\]

    C)  \[y\]

    D)  \[x\]

    Correct Answer: A

    Solution :

    Given,  \[y=1+\frac{x}{1!}+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+....\infty ={{e}^{x}}\] On differentiation w, r. t. to x. We get \[\frac{dy}{dx}={{e}^{x}}\]


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