Deriv of xtanx

  • #1
PrudensOptimus
641
0
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.
 

Answers and Replies

  • #2
mathman
Science Advisor
8,077
547
= ... + (2secx + sec^2(x))
First term should have d/dx(secx)=tanx.secx multipled, i.e. first term is 2sec2x.tanx.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
43,021
970
= ... + (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!
 
  • #4
PrudensOptimus
641
0
I found the answer this morning,

something like

2sec^2 x(tanx + 1)
 

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