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If y = tan ^-1 (secx + tan x). Find d^2y/dx^2 at x = π/4

If y = tan ^-1 (secx + tan x). Find
d^2y/dx^2 at x = π/4
 

Grade:11

2 Answers

Arun
25763 Points
3 years ago
Dear student
 
dy/dx = [1 /1 + (secx + tanx)² ][secx tanx + sec²x]
dy/ dx  = sec x(secx + tanx) /(1+ sec² x + tan²x + 2secx tanx)
dy/dx = sec x(secx + tanx) / 2secx ( secx + tanx)
dy/dx = ½
Hence d²y /dx² = 0
Sounak Dutta
10 Points
3 years ago
Y=tan-¹(secx+tanx)Y=tan-¹((1+sinx)/cos)Y=tan-¹((sinx/2+cosx/2)/(cosx/2-sinx/2))Y=tan-¹(tan(pi/4+x/2)Y=pi/4+x/2dy/dx=1/2d²y/dx²=0

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