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Higher Order Derivatives — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSHigher Order Derivatives1 Mark
MCQ
If $\text{y}=\text{ax}^{\text{n+1}}+\text{bx}^{-\text{n}}$ Then $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2} =$
  • A
    $n(n - 1) y$
  • ✓
    $n(n + 1) y$
  • C
    $ny$
  • D
    $n^2y$
✓

Answer

Correct option: B.
$n(n + 1) y$
Here
$\text{y}=\text{ax}^{\text{n}+1}+\text{bx}^{\text{-n}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{a}(\text{n}+1)\text{x}^\text{n}-\text{bn}\text{x}^{-\text{n}-1}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}$
$\therefore\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{x}^2\{\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}\}$
$=\text{n}(\text{n}+1)(\text{ax}^{\text{n}+1}+\text{b x}^{-\text{n}})$
$=\text{n}(\text{n}+1)\text{y}$

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