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

  • question_answer
    If \[y=f\left( \frac{2x-1}{{{x}^{2}}+1} \right)\]and \[{f}'(x)=\sin {{x}^{2}},\]then \[\frac{dy}{dx}=\] [IIT 1982]

    A) \[\frac{6{{x}^{2}}-2x+2}{{{({{x}^{2}}+1)}^{2}}}\sin {{\left( \frac{2x-1}{{{x}^{2}}+1} \right)}^{2}}\]

    B) \[\frac{6{{x}^{2}}-2x+2}{{{({{x}^{2}}+1)}^{2}}}{{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right)\]

    C) \[\frac{-2{{x}^{2}}+2x+2}{{{({{x}^{2}}+1)}^{2}}}{{\sin }^{2}}\left( \frac{2x-1}{{{x}^{2}}+1} \right)\] 

    D) \[\frac{-2{{x}^{2}}+2x+2}{{{({{x}^{2}}+1)}^{2}}}\sin {{\left( \frac{2x-1}{{{x}^{2}}+1} \right)}^{2}}\]

    Correct Answer: D

    Solution :

    • Let\[t=\frac{2x-1}{{{x}^{2}}+1},\] then \[\frac{dy}{dx}=f'(t).\frac{dt}{dx}\]                
    • \[=\sin {{t}^{2}}\frac{d}{dx}\left( \frac{2x-1}{{{x}^{2}}+1} \right)=\frac{2(1+x-{{x}^{2}})}{{{(1+{{x}^{2}})}^{2}}}.\sin {{\left( \frac{2x-1}{{{x}^{2}}+1} \right)}^{2}}\].


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