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

  • question_answer
    If \[{{y}^{2}}=p(x)\]is a polynomial of degree three, then \[2\frac{d}{dx}\left\{ {{y}^{3}}.\frac{{{d}^{2}}y}{d{{x}^{2}}} \right\}\]= [IIT 1988; RPET 2000]

    A) \[{p}'''(x)+p'(x)\]

    B) \[{p}''(x).{p}'''(x)\]

    C) \[p(x).{p}'''(x)\]

    D) Constant

    Correct Answer: C

    Solution :

    • \[2y\frac{dy}{dx}=p'(x)\Rightarrow 2\frac{dy}{dx}=\frac{{p}'(x)}{y}\]\[\Rightarrow 2\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{y{p}''(x)-{p}'(x){y}'}{{{y}^{2}}}\]             
    • Þ \[2{{y}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}}={{y}^{2}}{p}''(x)-y\frac{dy}{dx}{p}'(x)\]\[=p(x)\,{p}''(x)-\frac{1}{2}{{\{{p}'(x)\}}^{2}}\]                   
    • Þ \[2\frac{d}{dx}\left( {{y}^{3}}\frac{{{d}^{2}}y}{d{{x}^{2}}} \right)={p}'(x){p}''(x)+p(x){p}'''(x)-{p}'(x){p}''(x)\]                                                                        \[=p(x){p}'''(x)\].


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