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dangbeau
 one year ago
Find the derivative f(θ)= secθ^2
How to solve this?
dangbeau
 one year ago
Find the derivative f(θ)= secθ^2 How to solve this?

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

andrea408
 one year ago
Best ResponseYou've already chosen the best response.0f(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!

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

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

dangbeau
 one year ago
Best ResponseYou've already chosen the best response.0so 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?

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1the 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.

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

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

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

dangbeau
 one year ago
Best ResponseYou've already chosen the best response.0wait, 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

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

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1use 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)

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

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

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