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bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
y' = x*(Secx)^2+tanx
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
I found that y' is [x sec^2 x + tan x]
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
bb in 10 mins
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
I don't know how to do the next step...
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
y'' = x*2*Secx*(SecxTanx) + (Secx)^2 + (Secx)^2
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
simplify that
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
how did you arrive at that answer?
 2 years ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
you need the 2nd derivative
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
use product and power rules on x*(Secx)^2
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
derivative of (Secx)^2 is 2*Secx*(SecxTanx)
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
why wouldn't it be 2x (sec x tan x)
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
Is it because you have to use something called a chain rule?
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Sorry was away. Yes it is the chain rule and the product rule: derivative of u^a with respect to x is: a * (u)^(a1) * (du/dx)
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
Can you not do it with the chain rule it's long and tedious but..
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
u can't do it with any other rule
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
it's (Secx)^2 you have to use the chain rule.
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
x IS NOT a constant......................
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
neither is (Secx)^2
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
pg 91 number 27
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
ohhhhhhhhhhhhh
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
sorry, should have put that up..
 2 years ago

Freyes Group TitleBest ResponseYou've already chosen the best response.0
*waits for answer*
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Man I'd kill them for making me use the product rule on that.. (Secx)^2 = Secx*Secx = Secx*(SecxTanx) + (SecxTanx)*Secx
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
now... uhhh...
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Ok... let's start over.. One last time... http://www.youtube.com/watch?v=0Ox3V1aI38
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
^joke.. anyway.. y' = x*(Secx)^2+tanx y'' = x*[(Secx)^2] ' + 1*(Secx)^2 + (Secx)^2
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
I got lost
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
why is there two sec x ^2
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
There are 3 Secx^2.. which 2 are u talking about?
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
how did tan x become 2 (sec x ^2)
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Derivative of tangent is (Secx)^2.. you should memorize that. And derivative of cotangent is (Cscx)^2
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
yeah.. but you have 2 of sec ^2
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
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?
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Now look at the first term: {x*(Secx)^2}' = *using the product rule* = x*{(Secx)^2}' + {(Secx)^2}*1 right?
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
yes so sec (sec)' sec (sec)'
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
and sec' is euqal to sec*tan
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Not yet.. don't rush
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Now we need {(Secx)^2}' {(Secx)^2}' = {Secx*Secx}' = *using product rule* = Secx *(Secx)' + (Secx)'*Secx = = Secx * (SecxTanx) + (SecxTanx) * Secx
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Again.. how do we know that (Secx)' = SecxTanx.. you memorize it. (Cscx)' = CscxCotx btw
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
gots that
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
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
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Are you still following?
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
er... hang on for a sec
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
wouldn't it be... x sec sec (sec tan)(sec tan) ??
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
No.. Ok let's try Latex maybe then it's going to be easier to see.
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Did you follow how we got: \[{(Secx)^2}' = 2(Secx)^2(Tanx)\]
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Darn it the rest didn't show up
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
sec^2 = sec sec sec (sec tan) + sec (sec tan) sec^2 (sec tan) + sec^2 (sec tan) 2 sec^2 + 2 sec tan?
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
sec^2 = sec sec sec (sec tan) + sec (sec tan) <Here you're just multiplying it out. sec^2 (tan) + sec^2 (tan)
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
sec^2 (tan) + sec^2 (tan) = 2(sec^2 (tan))... something plus something is 2 of that something
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Did you follow?
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Okay.. THUS: \[(Sec^2x)'=2(Sec^2x)(Tanx)\]
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
yep and tan x > sec^2
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Now:.. uhmm i forgot what the question was.. lol
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
and.. on the constant multiple rule.. just 'add in' the extra x in the beginning?
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
do not forget that x is not a constant, x is a variable, so you have to use the product rule
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
\[2, \pi, e, 1/2, 0.005\] ^ Those are constants. x is a variable.
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
soo how would you .. take out the x?
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
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 \[g(x)=(Secx)^2\]
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
then.....
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
So \[(f(x)g(x))'=f(x)*g'(x)+f'(x)*g(x)\] or \[(x*(Secx)^2)'=x*(Sec^2x)'+x'*(Sec^2x)=x*2(Sec^2x)(Tanx)+1*(Sec^2x)\]
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Because earlier we showed that: \[{(Secx)^2}' = 2(Secx)^2(Tanx)\]
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
oh... that was the last step right? (cuz I THINK I got the answer)
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
No that's not the last step.. That's only the first term
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
. but then tan'> sec^2
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
WAIT wouldn't it be 0+sec x ^2
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
wait.. I got it nvrmind confused it with constant
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
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: \[(Tanx)'=Sec^2x\]
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
So the whole thing is: \[y''=2x(Sec^2x)(Tanx)+(Sec^2x)+(Sec^2x)\]
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
I got it! :D
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
and then you can put sec^2 sec^2 as 2(sec^2)
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
and that's the answer :) THANK YOU SO MUCH!!!!!!!!!!!!!!!!!!!!!
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
http://www.thehappydogspot.com/images/wrigleytheprayingcanineitsamiracle21605806.jpg
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
lol :) thank you!
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
np lol.. thanks for sticking to the question
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
I'm really in awe of your patience.. and of course, your skills
 2 years ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
thanks again!
 2 years ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
lol you're welcome.. just practice more
 2 years ago
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