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

  • question_answer
    If \[y={{\tan }^{-1}}\sqrt{\frac{1+\cos x}{1-\cos x}}\], then \[\frac{dy}{dx}\] is equal to [Roorkee 1995]

    A)            0

    B)            \[-\frac{1}{2}\]

    C)            ½

    D)            1

    Correct Answer: B

    Solution :

               \[y={{\tan }^{-1}}\sqrt{\frac{1+\cos x}{1-\cos x}}={{\tan }^{-1}}\sqrt{\frac{2{{\cos }^{2}}\frac{x}{2}}{2{{\sin }^{2}}\frac{x}{2}}}\]                 \[={{\tan }^{-1}}\cot \frac{x}{2}={{\tan }^{-1}}\tan \left( \frac{\pi }{2}-\frac{x}{2} \right)=\frac{\pi }{2}-\frac{x}{2}\]                    Þ  \[\frac{dy}{dx}=-\frac{1}{2}\].


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