## rainbow22 3 years ago find y" if y= x tan x

1. bahrom7893

y' = x*(Secx)^2+tanx

2. rainbow22

I found that y' is [x sec^2 x + tan x]

3. bahrom7893

bb in 10 mins

4. rainbow22

I don't know how to do the next step...

5. bahrom7893

y'' = x*2*Secx*(SecxTanx) + (Secx)^2 + (Secx)^2

6. bahrom7893

simplify that

7. rainbow22

how did you arrive at that answer?

8. ilikephysics2

you need the 2nd derivative

9. bahrom7893

use product and power rules on x*(Secx)^2

10. bahrom7893

derivative of (Secx)^2 is 2*Secx*(SecxTanx)

11. rainbow22

why wouldn't it be 2x (sec x tan x)

12. rainbow22

Is it because you have to use something called a chain rule?

13. bahrom7893

Sorry was away. Yes it is the chain rule and the product rule: derivative of u^a with respect to x is: a * (u)^(a-1) * (du/dx)

14. rainbow22

Can you not do it with the chain rule it's long and tedious but..

15. bahrom7893

u can't do it with any other rule

16. bahrom7893

it's (Secx)^2 you have to use the chain rule.

17. rainbow22

Oh, there was a way to use the constant multiple rule the sum/difference rule and the product rule to get the answer :( That's the way my book shows it. Because I'm learning the chain rule in the nxt ch.

18. bahrom7893

x IS NOT a constant......................

19. bahrom7893

neither is (Secx)^2

20. rainbow22

pg 91 number 27

21. bahrom7893

ohhhhhhhhhhhhh

22. rainbow22

sorry, should have put that up..

23. Freyes

24. bahrom7893

Man I'd kill them for making me use the product rule on that.. (Secx)^2 = Secx*Secx = Secx*(SecxTanx) + (SecxTanx)*Secx

25. rainbow22

now... uhhh...

26. bahrom7893

Ok... let's start over.. One last time... http://www.youtube.com/watch?v=0Ox3V-1aI38

27. bahrom7893

^joke.. anyway.. y' = x*(Secx)^2+tanx y'' = x*[(Secx)^2] ' + 1*(Secx)^2 + (Secx)^2

28. rainbow22

I got lost

29. bahrom7893

Where?

30. rainbow22

why is there two sec x ^2

31. bahrom7893

There are 3 Secx^2.. which 2 are u talking about?

32. rainbow22

how did tan x become 2 (sec x ^2)

33. bahrom7893

Derivative of tangent is (Secx)^2.. you should memorize that. And derivative of cotangent is -(Cscx)^2

34. rainbow22

yeah.. but you have 2 of sec ^2

35. bahrom7893

Okay... I was really hoping that we wouldn't start over again, but lets start over.. y' = x*(Secx)^2+tanx y '' then would be: {x*(Secx)^2}' + {tanx}' are you following so far?

36. rainbow22

yes

37. bahrom7893

Now look at the first term: {x*(Secx)^2}' = *using the product rule* = x*{(Secx)^2}' + {(Secx)^2}*1 right?

38. rainbow22

yes so sec (sec)' sec (sec)'

39. rainbow22

and sec' is euqal to sec*tan

40. bahrom7893

Not yet.. don't rush

41. rainbow22

okay

42. bahrom7893

Now we need {(Secx)^2}' {(Secx)^2}' = {Secx*Secx}' = *using product rule* = Secx *(Secx)' + (Secx)'*Secx = = Secx * (SecxTanx) + (SecxTanx) * Secx

43. bahrom7893

Again.. how do we know that (Secx)' = SecxTanx.. you memorize it. (Cscx)' = -CscxCotx btw

44. rainbow22

gots that

45. bahrom7893

So {(Secx)^2}' = 2(Secx)^2(Tanx) Plug that in here: {x*(Secx)^2}' = *using the product rule* = x*{(Secx)^2}' + {(Secx)^2}*1 = = x * 2(Secx)^2(Tanx) + (Secx)^2

46. bahrom7893

Are you still following?

47. rainbow22

er... hang on for a sec

48. bahrom7893

ok

49. rainbow22

wouldn't it be... x sec sec (sec tan)(sec tan) ??

50. bahrom7893

No.. Ok let's try Latex maybe then it's going to be easier to see.

51. bahrom7893

Did you follow how we got: ${(Secx)^2}' = 2(Secx)^2(Tanx)$

52. bahrom7893

Darn it the rest didn't show up

53. rainbow22

sec^2 = sec sec sec (sec tan) + sec (sec tan) sec^2 (sec tan) + sec^2 (sec tan) 2 sec^2 + 2 sec tan?

54. bahrom7893

sec^2 = sec sec sec (sec tan) + sec (sec tan) <---Here you're just multiplying it out. sec^2 (tan) + sec^2 (tan)

55. bahrom7893

sec^2 (tan) + sec^2 (tan) = 2(sec^2 (tan))... something plus something is 2 of that something

56. bahrom7893

Did you follow?

57. rainbow22

yes

58. bahrom7893

Okay.. THUS: $(Sec^2x)'=2(Sec^2x)(Tanx)$

59. rainbow22

yep and tan x -> sec^2

60. bahrom7893

Now:.. uhmm i forgot what the question was.. lol

61. rainbow22

and.. on the constant multiple rule.. just 'add in' the extra x in the beginning?

62. bahrom7893

do not forget that x is not a constant, x is a variable, so you have to use the product rule

63. rainbow22

uh.....?

64. bahrom7893

$2, \pi, e, 1/2, -0.005$ ^ Those are constants. x is a variable.

65. rainbow22

soo how would you .. take out the x?

66. bahrom7893

Not exactly. $y' = x*(Secx)^2+tanx$ Look at the first term: $x*(Secx)^2$that's just a product rule: $f(x)*g(x)$ where $f(x)=x$ and $g(x)=(Secx)^2$

67. rainbow22

then.....

68. bahrom7893

So $(f(x)g(x))'=f(x)*g'(x)+f'(x)*g(x)$ or $(x*(Secx)^2)'=x*(Sec^2x)'+x'*(Sec^2x)=x*2(Sec^2x)(Tanx)+1*(Sec^2x)$

69. bahrom7893

Because earlier we showed that: ${(Secx)^2}' = 2(Secx)^2(Tanx)$

70. rainbow22

oh... that was the last step right? (cuz I THINK I got the answer)

71. bahrom7893

No that's not the last step.. That's only the first term

72. rainbow22

-.- but then tan'-> sec^2

73. rainbow22

WAIT wouldn't it be 0+sec x ^2

74. rainbow22

wait.. I got it nvrmind confused it with constant

75. bahrom7893

Yes $y' = x*(Secx)^2+tanx$ The derivative of the first term was: $(x*Sec^2x)'=x*2(Sec^2x)(Tanx)+1*(Sec^2x)$ The derivative of the second term is: $(Tanx)'=Sec^2x$

76. bahrom7893

So the whole thing is: $y''=2x(Sec^2x)(Tanx)+(Sec^2x)+(Sec^2x)$

77. rainbow22

I got it! :D

78. rainbow22

and then you can put sec^2 sec^2 as 2(sec^2)

79. bahrom7893

YES!

80. rainbow22

and that's the answer :) THANK YOU SO MUCH!!!!!!!!!!!!!!!!!!!!!

81. bahrom7893
82. rainbow22

lol :) thank you!

83. bahrom7893

np lol.. thanks for sticking to the question

84. rainbow22

I'm really in awe of your patience.. and of course, your skills

85. rainbow22

thanks again!

86. bahrom7893

lol you're welcome.. just practice more

87. rainbow22

will do

88. rainbow22