A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Find the derivative f(θ)= secθ^2
How to solve this?
anonymous
 3 years ago
Find the derivative f(θ)= secθ^2 How to solve this?

This Question is Closed

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

anonymous
 3 years 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362873411456:dw dw:1362873506755:dw

anonymous
 3 years 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
 3 years 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.

anonymous
 3 years 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
 3 years 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
 3 years 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.

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how to get that answer?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yup, thanks sooo much ^^
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.