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Limits and Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits and Derivatives1 Mark
MCQ
If $\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}},$ then $\frac{\text{dy}}{\text{dx}}=$
  • A
    $-\frac{4\text{x}}{(\text{x}^2-1)^2}$
  • B
    $-\frac{4\text{x}}{\text{x}^2-1}$
  • C
    $\frac{1-\text{x}^2}{\text{4x}}$
  • D
    $\frac{4\text{x}}{\text{x}^2-1}$
✓

Answer

  1. $-\frac{4\text{x}}{(\text{x}^2-1)^2}$

Solution:

$\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}}$

$=\frac{\text{x}^2+1}{\text{x}^2-1}$

Differentiate both the sides with respect to x, we get

$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)}{(\text{x}^2-1)^2}$ (Quotient rule)

$=\frac{(\text{x}^2-1)(\text{2x}+0)-(\text{x}^2+1)(\text{2x}-0)}{(\text{x}^2-1)^2}$

$=\frac{\text{2x}^3-\text{2x}-\text{2x}^3-\text{2x}}{(\text{x}^2-1)^2}$

$=\frac{-\text{4x}}{(\text{x}^2-1)^2}$

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