What is the second, third, fourth and fifth derivative of tan(x)?

Question

14513 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-20T19:51:25

This is a possible way, I don't know if it is the easier o faster, but it is enjoying:

#y=tanx#

#y^((1))=1+tan^2x=1+y^2#

#y^((2))=2y*y^((1))=2y*(1+y^2)=2y+2y^3#

#y^((3))=2y^((1))+6y^2*y^((1))=2(1+y^2)+6y^2(1+y^2)=#

#=2+2y^2+6y^2+6y^4=2+8y^2+6y^4#

#y^((4))=16yy^((1))+24y^3y^((1))=16y(1+y^2)+24y^3(1+y^2)=#

#=16y+16y^3+24y^3+24y^5=16y+40y^3+24y^5#

#y^((5))=16y^((1))+120y^2y^((1))+120y^4y^((1))=#

#=16(1+y^2)+120y^2(1+y^2)+120y^4(1+y^2)=#

#=16+16y^2+120y^2+120y^4+120y^4+120y^6=#

#=16+136y^2+240y^4+120y^6=#

#=16+136tan^2x+240tan^4x+120tan^6x#.

I hope that it is correct, and... enjoying...