Second derivative of y = x tan x

[PrudensOptimus]

y = x tan x

y' = tan x + x*sec^2(x)

y'' = sec^2(x) + d/dx (x*sec^2(x))

= ... + (d/dx(sec^2(x)) + sec^2(x))

= ... + (2secx + sec^2(x))

= 2sec^3 x + sec^4(x)

what did I do wrong?? in book it says y'' = something else.

 

[mathman]

= ... + (2secx + sec^2(x))

First term should have d/dx(secx)=tanx.secx multipled, i.e. first term is 2sec²x.tanx.

 

[HallsofIvy]

= ... + (d/dx(sec^2(x)) + sec^2(x))

= ... + (2secx + sec^2(x))

= 2sec^3 x + sec^4(x)

You multiplied the first "sec^2 x" instead of adding!

 

[PrudensOptimus]

I found the answer this morning,

something like

2sec^2 x(tanx + 1)