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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    The derivative of \[{{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)\]with respect to \[{{\tan }^{-1}}\left( \frac{2x\sqrt{1-{{x}^{2}}}}{1-2{{x}^{2}}} \right)\]at \[x=0\], is

    A) \[\frac{1}{8}\]

    B) \[\frac{1}{4}\]

    C) \[\frac{1}{2}\]

    D) 1

    Correct Answer: B

    Solution :

    • Let \[y={{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)\], \[z={{\tan }^{-1}}\left( \frac{2x\,\sqrt{1-{{x}^{2}}}}{1-2{{x}^{2}}} \right)\]           
    • Putting \[x=\tan \theta ,\] we get           
    • \[y={{\tan }^{-1}}\left( \frac{\sec \theta -1}{\tan \theta } \right)={{\tan }^{-1}}\left( \tan \frac{\theta }{2} \right)=\frac{1}{2}{{\tan }^{-1}}x\]           
    • \[\therefore \frac{dy}{dx}=\frac{1}{2(1+{{x}^{2}})}\]           
    • Putting \[x=\sin \theta \], we get           
    • \[z={{\tan }^{-1}}\left( \frac{2\sin \theta \cos \theta }{\cos 2\theta } \right)={{\tan }^{-1}}(\tan 2\theta )=2\theta \]           
    • \[\therefore z=2{{\sin }^{-1}}x\Rightarrow \frac{dz}{dx}=\frac{2}{\sqrt{1-{{x}^{2}}}}\]           
    • Thus \[\frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\frac{1}{4(1+{{x}^{2}})}\sqrt{1-{{x}^{2}}}\Rightarrow {{\left( \frac{dy}{dx} \right)}_{x=0}}=\frac{1}{4}\].


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