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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

माना \(\mathrm{y}=\log _{\mathrm{e}}\left(\frac{1-\mathrm{x}^2}{1+\mathrm{x}^2}\right),-1<\mathrm{x}<1\) है। तो \(x=\frac{1}{2}\) पर \(225\left(y^{\prime}-y^{\prime \prime}\right)\) का मान ........... है।

  1. A \(732\)
  2. B \(746\)
  3. C \(742\)
  4. D \(736\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(736\)

Step-by-step Solution

Detailed explanation

\( y=\log _0\left(\frac{1-x^2}{1+x^2}\right) \) \( \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4}\) Again, \(\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}\) Again…
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