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

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

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

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

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

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

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

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

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

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

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.3Sorry 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)

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

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

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

rainbow22
 2 years ago
Best ResponseYou've already chosen the best response.2Oh, 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
 2 years ago
Best ResponseYou've already chosen the best response.3x IS NOT a constant......................

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

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

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

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

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

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

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

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

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

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.3Okay... 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?

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

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

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

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

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

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

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.3So {(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
 2 years ago
Best ResponseYou've already chosen the best response.3Are you still following?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.3Not 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 \[g(x)=(Secx)^2\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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