anonymous
  • anonymous
find y" if y= x tan x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
bahrom7893
  • bahrom7893
y' = x*(Secx)^2+tanx
anonymous
  • anonymous
I found that y' is [x sec^2 x + tan x]
bahrom7893
  • bahrom7893
bb in 10 mins

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anonymous
  • anonymous
I don't know how to do the next step...
bahrom7893
  • bahrom7893
y'' = x*2*Secx*(SecxTanx) + (Secx)^2 + (Secx)^2
bahrom7893
  • bahrom7893
simplify that
anonymous
  • anonymous
how did you arrive at that answer?
anonymous
  • anonymous
you need the 2nd derivative
bahrom7893
  • bahrom7893
use product and power rules on x*(Secx)^2
bahrom7893
  • bahrom7893
derivative of (Secx)^2 is 2*Secx*(SecxTanx)
anonymous
  • anonymous
why wouldn't it be 2x (sec x tan x)
anonymous
  • anonymous
Is it because you have to use something called a chain rule?
bahrom7893
  • 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
  • anonymous
Can you not do it with the chain rule it's long and tedious but..
bahrom7893
  • bahrom7893
u can't do it with any other rule
bahrom7893
  • bahrom7893
it's (Secx)^2 you have to use the chain rule.
anonymous
  • 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
  • bahrom7893
x IS NOT a constant......................
bahrom7893
  • bahrom7893
neither is (Secx)^2
anonymous
  • anonymous
pg 91 number 27
bahrom7893
  • bahrom7893
ohhhhhhhhhhhhh
anonymous
  • anonymous
sorry, should have put that up..
anonymous
  • anonymous
*waits for answer*
bahrom7893
  • bahrom7893
Man I'd kill them for making me use the product rule on that.. (Secx)^2 = Secx*Secx = Secx*(SecxTanx) + (SecxTanx)*Secx
anonymous
  • anonymous
now... uhhh...
bahrom7893
  • bahrom7893
Ok... let's start over.. One last time... http://www.youtube.com/watch?v=0Ox3V-1aI38
bahrom7893
  • bahrom7893
^joke.. anyway.. y' = x*(Secx)^2+tanx y'' = x*[(Secx)^2] ' + 1*(Secx)^2 + (Secx)^2
anonymous
  • anonymous
I got lost
bahrom7893
  • bahrom7893
Where?
anonymous
  • anonymous
why is there two sec x ^2
bahrom7893
  • bahrom7893
There are 3 Secx^2.. which 2 are u talking about?
anonymous
  • anonymous
how did tan x become 2 (sec x ^2)
bahrom7893
  • bahrom7893
Derivative of tangent is (Secx)^2.. you should memorize that. And derivative of cotangent is -(Cscx)^2
anonymous
  • anonymous
yeah.. but you have 2 of sec ^2
bahrom7893
  • 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
  • anonymous
yes
bahrom7893
  • bahrom7893
Now look at the first term: {x*(Secx)^2}' = *using the product rule* = x*{(Secx)^2}' + {(Secx)^2}*1 right?
anonymous
  • anonymous
yes so sec (sec)' sec (sec)'
anonymous
  • anonymous
and sec' is euqal to sec*tan
bahrom7893
  • bahrom7893
Not yet.. don't rush
anonymous
  • anonymous
okay
bahrom7893
  • bahrom7893
Now we need {(Secx)^2}' {(Secx)^2}' = {Secx*Secx}' = *using product rule* = Secx *(Secx)' + (Secx)'*Secx = = Secx * (SecxTanx) + (SecxTanx) * Secx
bahrom7893
  • bahrom7893
Again.. how do we know that (Secx)' = SecxTanx.. you memorize it. (Cscx)' = -CscxCotx btw
anonymous
  • anonymous
gots that
bahrom7893
  • 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
  • bahrom7893
Are you still following?
anonymous
  • anonymous
er... hang on for a sec
bahrom7893
  • bahrom7893
ok
anonymous
  • anonymous
wouldn't it be... x sec sec (sec tan)(sec tan) ??
bahrom7893
  • bahrom7893
No.. Ok let's try Latex maybe then it's going to be easier to see.
bahrom7893
  • bahrom7893
Did you follow how we got: \[{(Secx)^2}' = 2(Secx)^2(Tanx)\]
bahrom7893
  • bahrom7893
Darn it the rest didn't show up
anonymous
  • 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
  • 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
  • bahrom7893
sec^2 (tan) + sec^2 (tan) = 2(sec^2 (tan))... something plus something is 2 of that something
bahrom7893
  • bahrom7893
Did you follow?
anonymous
  • anonymous
yes
bahrom7893
  • bahrom7893
Okay.. THUS: \[(Sec^2x)'=2(Sec^2x)(Tanx)\]
anonymous
  • anonymous
yep and tan x -> sec^2
bahrom7893
  • bahrom7893
Now:.. uhmm i forgot what the question was.. lol
anonymous
  • anonymous
and.. on the constant multiple rule.. just 'add in' the extra x in the beginning?
bahrom7893
  • bahrom7893
do not forget that x is not a constant, x is a variable, so you have to use the product rule
anonymous
  • anonymous
uh.....?
bahrom7893
  • bahrom7893
\[2, \pi, e, 1/2, -0.005\] ^ Those are constants. x is a variable.
anonymous
  • anonymous
soo how would you .. take out the x?
bahrom7893
  • 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
  • anonymous
then.....
bahrom7893
  • 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
  • bahrom7893
Because earlier we showed that: \[{(Secx)^2}' = 2(Secx)^2(Tanx)\]
anonymous
  • anonymous
oh... that was the last step right? (cuz I THINK I got the answer)
bahrom7893
  • bahrom7893
No that's not the last step.. That's only the first term
anonymous
  • anonymous
-.- but then tan'-> sec^2
anonymous
  • anonymous
WAIT wouldn't it be 0+sec x ^2
anonymous
  • anonymous
wait.. I got it nvrmind confused it with constant
bahrom7893
  • 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
  • bahrom7893
So the whole thing is: \[y''=2x(Sec^2x)(Tanx)+(Sec^2x)+(Sec^2x)\]
anonymous
  • anonymous
I got it! :D
anonymous
  • anonymous
and then you can put sec^2 sec^2 as 2(sec^2)
bahrom7893
  • bahrom7893
YES!
anonymous
  • anonymous
and that's the answer :) THANK YOU SO MUCH!!!!!!!!!!!!!!!!!!!!!
bahrom7893
  • bahrom7893
http://www.the-happy-dog-spot.com/images/wrigley-the-praying-canine-its-a-miracle-21605806.jpg
anonymous
  • anonymous
lol :) thank you!
bahrom7893
  • bahrom7893
np lol.. thanks for sticking to the question
anonymous
  • anonymous
I'm really in awe of your patience.. and of course, your skills
anonymous
  • anonymous
thanks again!
bahrom7893
  • bahrom7893
lol you're welcome.. just practice more
anonymous
  • anonymous
will do
anonymous
  • anonymous
adios.

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