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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2000

  • question_answer
    The derivative of \[{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] w.r.t.\[{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\]is:

    A)  \[-1\]                                   

    B)  \[1\]

    C)  2                                            

    D)  4

    Correct Answer: B

    Solution :

    \[u={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right),\upsilon =co{{s}^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] put \[x=\tan \theta \,x=\tan \theta \] \[u=2\theta \]                   \[u=2\theta \] \[\frac{du}{d\theta }=1\]                             \[\frac{du}{d\theta }=1\] \[\frac{du}{d\upsilon }=1\]


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