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find y" if y= x tan x

Mathematics
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y' = x*(Secx)^2+tanx
I found that y' is [x sec^2 x + tan x]
bb in 10 mins

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Other answers:

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

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