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Class 12 Continuity And Differentiability · Long Answer3 Markssecond_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #146

For y=ln tan(π/4+x/2), show y′=sec x and find y″ at x=π/4, on intervals where tan(π/4+x/2)>0.

✅ Step-by-Step Solution

Final Answer: y′=sec x; y″(π/4)=√2.
  1. Put u=π/4+x/2. By the chain rule y′=sec²u/(2tan u)=1/(2sin u cos u)=1/sin2u.
  2. Since 2u=π/2+x, sin2u=cos x, so y′=sec x.
  3. Then y″=sec x tan x. At π/4, sec x=√2 and tan x=1.
  4. Thus y′=sec x and y″(π/4)=√2.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Put u=π/4+x/2. By the chain rule y′=sec²u/(2tan u)=1/(2sin u cos u)=1/sin2u.
  • [1 mark] Since 2u=π/2+x, sin2u=cos x, so y′=sec x.
  • [1 mark] Then y″=sec x tan x. At π/4, sec x=√2 and tan x=1.
  • [0.5 marks] Thus y′=sec x and y″(π/4)=√2.

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