JEE Main & Advanced Mathematics Differentiation Question Bank Differentiation by Substitution

  • question_answer
    If \[y=\sin px\] and \[{{y}_{n}}\] is the nth derivative of y, then \[\left| \,\begin{matrix}    y & {{y}_{1}} & {{y}_{2}}  \\    {{y}_{3}} & {{y}_{4}} & {{y}_{5}}  \\    {{y}_{6}} & {{y}_{7}} & {{y}_{8}}  \\ \end{matrix}\, \right|\] is equal to [AMU 2002]

    A)             1

    B)             0

    C)            ? 1

    D)            None of these

    Correct Answer: B

    Solution :

               \[D=\left| \,\begin{matrix}    \sin px & p\cos px & -{{p}^{2}}\sin px  \\    -{{p}^{3}}\cos px & {{p}^{4}}\sin px & {{p}^{5}}\cos px  \\    -{{p}^{6}}\sin px & -{{p}^{7}}\cos px & {{p}^{8}}\sin px  \\ \end{matrix}\, \right|\]              \[={{p}^{9}}\left| \,\begin{matrix}    \sin px & p\cos px & -{{p}^{2}}\sin px  \\    -\cos px & p\sin px & {{p}^{2}}\cos px  \\    -\sin px & -p\cos px & {{p}^{2}}\sin px  \\ \end{matrix}\, \right|\]                     \[=-{{p}^{9}}\,\left| \,\begin{matrix}    \sin px & p\cos px & -{{p}^{2}}\sin px  \\    \cos px & p\sin px & {{p}^{2}}\cos px  \\    \sin px & p\cos px & -{{p}^{2}}\sin px  \\ \end{matrix}\, \right|=0.\]


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