find y" if y= x tan x

- anonymous

find y" if y= x tan x

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- bahrom7893

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

- anonymous

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

- bahrom7893

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## More answers

- anonymous

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

- bahrom7893

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

- bahrom7893

simplify that

- anonymous

how did you arrive at that answer?

- anonymous

you need the 2nd derivative

- bahrom7893

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

- bahrom7893

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

- anonymous

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

- anonymous

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

- 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)

- anonymous

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

- bahrom7893

u can't do it with any other rule

- bahrom7893

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

- anonymous

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.

- bahrom7893

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

- bahrom7893

neither is (Secx)^2

- anonymous

pg 91 number 27

##### 1 Attachment

- bahrom7893

ohhhhhhhhhhhhh

- anonymous

sorry, should have put that up..

- anonymous

*waits for answer*

- bahrom7893

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

- anonymous

now... uhhh...

- bahrom7893

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

- bahrom7893

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

- anonymous

I got lost

- bahrom7893

Where?

- anonymous

why is there two sec x ^2

- bahrom7893

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

- anonymous

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

- bahrom7893

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

- anonymous

yeah.. but you have 2 of sec ^2

- 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?

- anonymous

yes

- bahrom7893

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

- anonymous

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

- anonymous

and sec' is euqal to sec*tan

- bahrom7893

Not yet.. don't rush

- anonymous

okay

- bahrom7893

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

- bahrom7893

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

- anonymous

gots that

- 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

- bahrom7893

Are you still following?

- anonymous

er... hang on for a sec

- bahrom7893

ok

- anonymous

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

- bahrom7893

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

- bahrom7893

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

- bahrom7893

Darn it the rest didn't show up

- anonymous

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

- bahrom7893

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

- bahrom7893

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

- bahrom7893

Did you follow?

- anonymous

yes

- bahrom7893

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

- anonymous

yep
and tan x -> sec^2

- bahrom7893

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

- anonymous

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

- bahrom7893

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

- anonymous

uh.....?

- bahrom7893

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

- anonymous

soo how would you .. take out the x?

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

- anonymous

then.....

- bahrom7893

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

- bahrom7893

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

- anonymous

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

- bahrom7893

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

- anonymous

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

- anonymous

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

- anonymous

wait.. I got it nvrmind
confused it with constant

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

- bahrom7893

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

- anonymous

I got it! :D

- anonymous

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

- bahrom7893

YES!

- anonymous

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

- bahrom7893

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

- anonymous

lol :) thank you!

- bahrom7893

np lol.. thanks for sticking to the question

- anonymous

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

- anonymous

thanks again!

- bahrom7893

lol you're welcome.. just practice more

- anonymous

will do

- anonymous

adios.

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