JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(x = 3\,tan\,t\) and \(y = 3\,sec\,t,\) then the value of \(\frac{{{d^2}y}}{{d{x^2}}}\) at \(t = \frac {\pi }{4},\) is
- A \(\frac{3}{{2\sqrt 2 }}\)
- B \(\frac{1}{{3\sqrt 2 }}\)
- C \(\frac {1}{6}\)
- D \(\frac{1}{{6\sqrt 2 }}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{{6\sqrt 2 }}\)
Step-by-step Solution
Detailed explanation
\(\frac{{dx}}{{dt}} = 3{\sec ^2}t\) \(\frac{{dy}}{{dt}} = 3\sec t\,\tan t\) \(\frac{{dy}}{{dx}} = \frac{{\tan t}}{{\sec t}} = \sin t\) \(\frac{{{d^2}y}}{{d{x^2}}} = \cos \,t\frac{{dt}}{{dx}}\)…
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