Leibniz Rule MCQ

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• If y(x) = tan-1x then
  ক

  (yn+1)(0) = (n-1)(yn-1)0

  খ

  (yn+1)(0) = n(n-1)(yn-1)0

  গ

  (yn+1)(0) = -(n-1)(yn-1)0

  ঘ

  (yn+1)(0) = -n(n-1)(yn-1)0
• The nth derivative of x2y2 + (1-x2)y1 + xy = 0 is,
  ক

  x2yn+2 + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0

  খ

  x2yn+2 + yn+1(2nx-x2) + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0

  গ

  x2yn+2 + yn+1(2nx+1-x2) + yn(2n2-2n-2nx+x)-yn – 1(2n2-3n)=0

  ঘ

  x2yn+2 + yn+1(2nx+1-x2) + xyn2n2-yn – 1(2n2-3n)=0
• Find nth derivative of x2y2 + xy1 + y = 0
  ক

  x2yn+2 + (2n+1)xyn+1 + (n2+1)yn = 0

  খ

  x2yn+2 + nxyn+1 + (n2+1)yn = 0

  গ

  x2yn+2 + (2n+1)xyn+1 + n2yn = 0

  ঘ

  x2yn+2 + 2nxyn+1 + n2yn = 0
• If nth derivative of xy3 + x2y2 + x3y0 = 0 then order of its nth differential equation is,
  ক

  n

  খ

  n+1

  গ

  n+2

  ঘ

  n+3