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Model Paper 10 — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 101 Mark
MCQ
If $y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}$ then $\frac{d y}{d x}$ is equal to
  • A
    $\frac{-4 x}{x^2-1}$
  • B
    $\frac{1-x^2}{4 x}$
  • C
    $\frac{-4 x}{\left(x^2-1\right)^2}$
  • D
    $\frac{4 x}{x^2-1}$
✓

Answer

(c) $\frac{-4 x}{\left(x^2-1\right)^2}$
Explanation: Given $y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{z^2}} \Rightarrow y=\frac{x^2+1}{x^2-1}$
$\therefore \quad \frac{d y}{d x}=\frac{\left(x^2-1\right) \cdot 2 x-\left(x^2+1\right) \cdot 2 x}{\left(x^2-1\right)^2}$
$=\frac{2 x\left(x^2-1-x^2-1\right)}{\left(x^2-1\right)^2}=\frac{2 x(-2)}{\left(x^2-1\right)^2}=\frac{-4 x}{\left(x^2-1\right)^2}$

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