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JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    If \[s=\sqrt{{{t}^{2}}+1}\], then \[\frac{{{d}^{2}}s}{d{{t}^{2}}}\] is equal to

    A) \[\frac{1}{s}\]

    B) \[\frac{1}{{{s}^{2}}}\]

    C) \[\frac{1}{{{s}^{3}}}\]

    D) \[\frac{1}{{{s}^{4}}}\]

    Correct Answer: C

    Solution :

    [c] \[s=\sqrt{{{t}^{2}}+1}\] \[\Rightarrow \frac{ds}{dt}=\frac{t}{\sqrt{{{t}^{2}}+1}}\Rightarrow \frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{1}{\sqrt{{{\left( {{t}^{2}}+1 \right)}^{3}}}}\] \[\Rightarrow \frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{1}{{{s}^{3}}}\]


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