Here's the question you clicked on:
dangbeau
Find the derivative f(θ)= secθ^2 How to solve this?
d/dx f(g(x))=f'(g(x))g'(x)
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!
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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?
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.
hm... i have another answer \[\sec ^{2} \theta=\sec \theta ^{2}\] it's equal, right?
no, of course not. sec^2(th)= sec th*sec th sec(th^2)=sec(th*th)
No, it's not correct.Re-write sec^2 x as (sec x)^2, then power and chain rules.
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
how to get that answer?
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)
okay - so the chain rule makes more sense to you now?
yup, thanks sooo much ^^