rainbow22
find y" if y= x tan x
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bahrom7893
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y' = x*(Secx)^2+tanx
rainbow22
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I found that y' is [x sec^2 x + tan x]
bahrom7893
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bb in 10 mins
rainbow22
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I don't know how to do the next step...
bahrom7893
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y'' = x*2*Secx*(SecxTanx) + (Secx)^2 + (Secx)^2
bahrom7893
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simplify that
rainbow22
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how did you arrive at that answer?
ilikephysics2
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you need the 2nd derivative
bahrom7893
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use product and power rules on x*(Secx)^2
bahrom7893
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derivative of (Secx)^2 is 2*Secx*(SecxTanx)
rainbow22
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why wouldn't it be 2x (sec x tan x)
rainbow22
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Is it because you have to use something called a chain rule?
bahrom7893
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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)
rainbow22
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Can you not do it with the chain rule
it's long and tedious but..
bahrom7893
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u can't do it with any other rule
bahrom7893
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it's (Secx)^2 you have to use the chain rule.
rainbow22
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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
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x IS NOT a constant......................
bahrom7893
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neither is (Secx)^2
rainbow22
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pg 91 number 27
bahrom7893
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ohhhhhhhhhhhhh
rainbow22
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sorry, should have put that up..
Freyes
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*waits for answer*
bahrom7893
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Man I'd kill them for making me use the product rule on that..
(Secx)^2 = Secx*Secx = Secx*(SecxTanx) + (SecxTanx)*Secx
rainbow22
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now... uhhh...
bahrom7893
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^joke.. anyway..
y' = x*(Secx)^2+tanx
y'' = x*[(Secx)^2] ' + 1*(Secx)^2 + (Secx)^2
rainbow22
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I got lost
bahrom7893
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Where?
rainbow22
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why is there two sec x ^2
bahrom7893
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There are 3 Secx^2.. which 2 are u talking about?
rainbow22
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how did tan x become 2 (sec x ^2)
bahrom7893
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Derivative of tangent is (Secx)^2.. you should memorize that. And derivative of cotangent is -(Cscx)^2
rainbow22
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yeah.. but you have 2 of sec ^2
bahrom7893
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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?
rainbow22
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yes
bahrom7893
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Now look at the first term:
{x*(Secx)^2}' = *using the product rule* = x*{(Secx)^2}' + {(Secx)^2}*1 right?
rainbow22
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yes
so
sec (sec)' sec (sec)'
rainbow22
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and sec' is euqal to sec*tan
bahrom7893
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Not yet.. don't rush
rainbow22
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okay
bahrom7893
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Now we need {(Secx)^2}'
{(Secx)^2}' = {Secx*Secx}' = *using product rule* = Secx *(Secx)' + (Secx)'*Secx =
= Secx * (SecxTanx) + (SecxTanx) * Secx
bahrom7893
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Again.. how do we know that (Secx)' = SecxTanx.. you memorize it. (Cscx)' = -CscxCotx btw
rainbow22
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gots that
bahrom7893
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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
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Are you still following?
rainbow22
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er... hang on for a sec
bahrom7893
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ok
rainbow22
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wouldn't it be... x sec sec (sec tan)(sec tan) ??
bahrom7893
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No.. Ok let's try Latex maybe then it's going to be easier to see.
bahrom7893
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Did you follow how we got:
\[{(Secx)^2}' = 2(Secx)^2(Tanx)\]
bahrom7893
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Darn it the rest didn't show up
rainbow22
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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
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sec^2 = sec sec
sec (sec tan) + sec (sec tan) <---Here you're just multiplying it out.
sec^2 (tan) + sec^2 (tan)
bahrom7893
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sec^2 (tan) + sec^2 (tan) = 2(sec^2 (tan))... something plus something is 2 of that something
bahrom7893
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Did you follow?
rainbow22
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yes
bahrom7893
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Okay.. THUS:
\[(Sec^2x)'=2(Sec^2x)(Tanx)\]
rainbow22
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yep
and tan x -> sec^2
bahrom7893
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Now:.. uhmm i forgot what the question was.. lol
rainbow22
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and.. on the constant multiple rule.. just 'add in' the extra x in the beginning?
bahrom7893
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do not forget that x is not a constant, x is a variable, so you have to use the product rule
rainbow22
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uh.....?
bahrom7893
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\[2, \pi, e, 1/2, -0.005\] ^ Those are constants. x is a variable.
rainbow22
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soo how would you .. take out the x?
bahrom7893
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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
\[g(x)=(Secx)^2\]
rainbow22
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then.....
bahrom7893
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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
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Because earlier we showed that:
\[{(Secx)^2}' = 2(Secx)^2(Tanx)\]
rainbow22
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oh...
that was the last step right?
(cuz I THINK I got the answer)
bahrom7893
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No that's not the last step.. That's only the first term
rainbow22
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-.- but then tan'-> sec^2
rainbow22
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WAIT
wouldn't it be 0+sec x ^2
rainbow22
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wait.. I got it nvrmind
confused it with constant
bahrom7893
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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
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So the whole thing is:
\[y''=2x(Sec^2x)(Tanx)+(Sec^2x)+(Sec^2x)\]
rainbow22
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I got it! :D
rainbow22
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and then you can put sec^2 sec^2
as 2(sec^2)
bahrom7893
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YES!
rainbow22
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and that's the answer :)
THANK YOU SO MUCH!!!!!!!!!!!!!!!!!!!!!
rainbow22
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lol :) thank you!
bahrom7893
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np lol.. thanks for sticking to the question
rainbow22
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I'm really in awe of your patience.. and of course, your skills
rainbow22
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thanks again!
bahrom7893
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lol you're welcome.. just practice more
rainbow22
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will do
rainbow22
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adios.