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inkyvoydBest ResponseYou've already chosen the best response.1
d/dx f(g(x))=f'(g(x))g'(x)
 one year ago

andrea408Best ResponseYou've already chosen the best response.0
f(θ) = y f'(θ) = dy/dθ u = sec(θ) du/dθ = sec(θ)tan(θ) y = u² dy/du = 2u Then, f'(θ) = 2usec(θ)tan(θ) ==> 2(sec²(θ))sec(θ)tan(θ) ==> 2sec³(θ)tan(θ) I hope this helps!
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
f(x)=x^2 g(th)=sec th
 one year ago

kausarsalleyBest ResponseYou've already chosen the best response.0
dw:1362873411456:dw dw:1362873506755:dw
 one year ago

dangbeauBest ResponseYou've already chosen the best response.0
so the answer is the same as me f'(θ)= \[2\sec ^{2}\theta \tan \theta\], right? but i don't know why the right answer is \[2\theta \sec \theta ^{2} \tan \theta ^{2}\] can anyone explain it?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
the answer is the one that you got and others gave. http://www.wolframalpha.com/input/?i=d%2Fdx+sec^2+x the answer 2th sec^2 th tan^2 th is incorrect.
 one year ago

dangbeauBest ResponseYou've already chosen the best response.0
hm... i have another answer \[\sec ^{2} \theta=\sec \theta ^{2}\] it's equal, right?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
no, of course not. sec^2(th)= sec th*sec th sec(th^2)=sec(th*th)
 one year ago

NoelGrecoBest ResponseYou've already chosen the best response.0
No, it's not correct.Rewrite sec^2 x as (sec x)^2, then power and chain rules.
 one year ago

dangbeauBest ResponseYou've already chosen the best response.0
wait, this one http://www.wolframalpha.com/input/?i=d%2Fdx+sec^2+x you solve it as f(x)=sec^2x, but i ask f(x)=secx^2 o.0
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
oh  http://www.wolframalpha.com/input/?i=d%2Fdx+sec+x^2
 one year ago

dangbeauBest ResponseYou've already chosen the best response.0
how to get that answer?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
use f(x)=sec x g(x)=x^2 then f'(g(x))g'(x)=tan(x^2)sec(x^2)*d/dx x^2=2x tan(x^2)sec(x^2)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
does that make sense?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
okay  so the chain rule makes more sense to you now?
 one year ago

dangbeauBest ResponseYou've already chosen the best response.0
yup, thanks sooo much ^^
 one year ago
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