JEE Main & Advanced Mathematics Differentiation Question Bank Differentiation by Substitution

  • question_answer
    If \[y={{x}^{2}}{{e}^{mx}}\], where m is a constant, then \[\frac{{{d}^{3}}y}{d{{x}^{3}}}=\] [MP PET 1987]

    A)            \[m{{e}^{mx}}({{m}^{2}}{{x}^{2}}+6mx+6)\]

    B)            \[2{{m}^{3}}x{{e}^{mx}}\]

    C)            \[m{{e}^{mx}}({{m}^{2}}{{x}^{2}}+2mx+2)\]

    D)            None of these

    Correct Answer: A

    Solution :

               \[y={{x}^{2}}{{e}^{mx}}\]            Differentiating w.r.t. \[x\], we get \[\frac{dy}{dx}=2x{{e}^{mx}}+m{{x}^{2}}{{e}^{mx}}\]            Again, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2({{e}^{mx}}+mx{{e}^{mx}})+m(2x{{e}^{mx}}+{{x}^{2}}m{{e}^{mx}})\]            or \[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{mx}}({{m}^{2}}{{x}^{2}}+4mx+2)\]            Again, \[\frac{{{d}^{3}}y}{d{{x}^{3}}}={{e}^{mx}}[{{m}^{3}}{{x}^{2}}+4{{m}^{2}}x+2m+2{{m}^{2}}x+4m]\]                                       \[={{e}^{mx}}[{{m}^{3}}{{x}^{2}}+6{{m}^{2}}x+6m]\]                    \[\Rightarrow \frac{{{d}^{3}}y}{d{{x}^{3}}}=m{{e}^{mx}}({{m}^{2}}{{x}^{2}}+6mx+6)\].


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