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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
ohhhhhhhhhhhhh
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.3
Darn it the rest didn't show up
 one year 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?
 one year 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)
 one year 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
 one year ago

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

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

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

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

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
uh.....?
 one year 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.
 one year ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
soo how would you .. take out the x?
 one year 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 \[g(x)=(Secx)^2\]
 one year ago

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
then.....
 one year 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 \[(x*(Secx)^2)'=x*(Sec^2x)'+x'*(Sec^2x)=x*2(Sec^2x)(Tanx)+1*(Sec^2x)\]
 one year ago

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

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

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

rainbow22 Group TitleBest ResponseYou've already chosen the best response.2
wait.. I got it nvrmind confused it with constant
 one year 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: \[(Tanx)'=Sec^2x\]
 one year ago

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

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

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

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

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

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

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

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