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

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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:

$y = tan x$ $y=tanx$

$y^{(1)} = 1 + {tan}^{2} x = 1 + y^{2}$ $y^((1))=1+tan^2x=1+y^2$

$y^{(2)} = 2 y \cdot y^{(1)} = 2 y \cdot (1 + y^{2}) = 2 y + 2 y^{3}$ $y^((2))=2y*y^((1))=2y*(1+y^2)=2y+2y^3$

$y^{(3)} = 2 y^{(1)} + 6 y^{2} \cdot y^{(1)} = 2 (1 + y^{2}) + 6 y^{2} (1 + y^{2}) =$ $y^((3))=2y^((1))+6y^2*y^((1))=2(1+y^2)+6y^2(1+y^2)=$

$= 2 + 2 y^{2} + 6 y^{2} + 6 y^{4} = 2 + 8 y^{2} + 6 y^{4}$ $=2+2y^2+6y^2+6y^4=2+8y^2+6y^4$

$y^{(4)} = 16 y y^{(1)} + 24 y^{3} y^{(1)} = 16 y (1 + y^{2}) + 24 y^{3} (1 + y^{2}) =$ $y^((4))=16yy^((1))+24y^3y^((1))=16y(1+y^2)+24y^3(1+y^2)=$

$= 16 y + 16 y^{3} + 24 y^{3} + 24 y^{5} = 16 y + 40 y^{3} + 24 y^{5}$ $=16y+16y^3+24y^3+24y^5=16y+40y^3+24y^5$

$y^{(5)} = 16 y^{(1)} + 120 y^{2} y^{(1)} + 120 y^{4} y^{(1)} =$ $y^((5))=16y^((1))+120y^2y^((1))+120y^4y^((1))=$

$= 16 (1 + y^{2}) + 120 y^{2} (1 + y^{2}) + 120 y^{4} (1 + y^{2}) =$ $=16(1+y^2)+120y^2(1+y^2)+120y^4(1+y^2)=$

$= 16 + 16 y^{2} + 120 y^{2} + 120 y^{4} + 120 y^{4} + 120 y^{6} =$ $=16+16y^2+120y^2+120y^4+120y^4+120y^6=$

$= 16 + 136 y^{2} + 240 y^{4} + 120 y^{6} =$ $=16+136y^2+240y^4+120y^6=$

$= 16 + 136 {tan}^{2} x + 240 {tan}^{4} x + 120 {tan}^{6} x$ $=16+136tan^2x+240tan^4x+120tan^6x$ .

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