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

  • question_answer
    If \[y=\frac{{{a}^{{{\cos }^{-1}}x}}}{1+{{a}^{{{\cos }^{-1}}x}}}\]and \[z={{a}^{{{\cos }^{-1}}x}}\],  then \[\frac{dy}{dx}\]= [MP PET 1994]

    A)            \[\frac{1}{1+{{a}^{{{\cos }^{-1}}x}}}\]

    B)            \[-\frac{1}{1+{{a}^{{{\cos }^{-1}}x}}}\]

    C)            \[\frac{1}{{{(1+{{a}^{{{\cos }^{-1}}x}})}^{2}}}\]

    D)            None of these

    Correct Answer: C

    Solution :

               \[y=\frac{{{a}^{{{\cos }^{-1}}x}}}{1+{{a}^{{{\cos }^{-1}}x}}},\,\,\,z={{a}^{{{\cos }^{-1}}x}}\] \[\Rightarrow y=\frac{z}{1+z}\]             \[\Rightarrow \frac{dy}{dz}=\frac{(1+z)1-z(1)}{{{(1+z)}^{2}}}=\frac{1}{{{(1+z)}^{2}}}\]\[=\frac{1}{{{(1+{{a}^{{{\cos }^{-1}}x}})}^{2}}}\].


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