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

If \(y = {\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}}\) , then \(\left( {{x^2} - 1} \right)\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}}\) is equal to

  1. A \(12\,y\)
  2. B \(224\,y^2\)
  3. C \(225\,y^2\)
  4. D \(225\,y\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(225\,y\)

Step-by-step Solution

Detailed explanation

\(y = {\left\{ {x + \sqrt {{x^2} - 1} } \right\}^{15}} + {\left\{ {x - \sqrt {{x^2} - 1} } \right\}^{15}}\) Differentiate w.r.t.\('x'\) \(\frac{{dy}}{{dx}} = 15{\left( {x + \sqrt {{x^2} - 1} } \right)^{14}}\left( {1 + \frac{x}{{\sqrt {{x^2} - 1} }}} \right) + \)…
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