Question

find y" if y= x tan x

Answer 1

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

Answer 2

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

Answer 3

bb in 10 mins

Answer 4

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

Answer 5

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

Answer 6

simplify that

Answer 7

how did you arrive at that answer?

Answer 8

you need the 2nd derivative

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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)

Answer 14

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

Answer 15

u can't do it with any other rule

Answer 16

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

Answer 17

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.

Answer 18

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

Answer 19

neither is (Secx)^2

Answer 20

pg 91 number 27

Answer 21

ohhhhhhhhhhhhh

Answer 22

sorry, should have put that up..

Answer 23

*waits for answer*

Answer 24

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

Answer 25

now... uhhh...

Answer 26

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

Answer 27

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

Answer 28

I got lost

Answer 29

Where?

Answer 30

why is there two sec x ^2

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

yeah.. but you have 2 of sec ^2

Answer 35

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?

Answer 36

yes

Answer 37

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

Answer 38

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

Answer 39

and sec' is euqal to sec*tan

Answer 40

Not yet.. don't rush

Answer 41

okay

Answer 42

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

Answer 43

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

Answer 44

gots that

Answer 45

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

Answer 46

Are you still following?

Answer 47

er... hang on for a sec

Answer 48

ok

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

Darn it the rest didn't show up

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

Did you follow?

Answer 57

yes

Answer 58

Okay.. THUS:
\[(Sec^2x)'=2(Sec^2x)(Tanx)\]

Answer 59

yep
and tan x -> sec^2

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

uh.....?

Answer 64

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

Answer 65

soo how would you .. take out the x?

Answer 66

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
\[g(x)=(Secx)^2\]

Answer 67

then.....

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

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

Answer 72

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

Answer 73

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

Answer 74

wait.. I got it nvrmind
confused it with constant

Answer 75

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:
\[(Tanx)'=Sec^2x\]

Answer 76

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

Answer 77

I got it! :D

Answer 78

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

Answer 79

YES!

Answer 80

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

Answer 81

http://www.the-happy-dog-spot.com/images/wrigley-the-praying-canine-its-a-miracle-21605806.jpg

Answer 82

lol :) thank you!

Answer 83

np lol.. thanks for sticking to the question

Answer 84

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

Answer 85

thanks again!

Answer 86

lol you're welcome.. just practice more

Answer 87

will do

Answer 88

adios.