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

1. inkyvoyd

use the chain rule

2. inkyvoyd

d/dx f(g(x))=f'(g(x))g'(x)

3. andrea408

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!

4. inkyvoyd

f(x)=x^2 g(th)=sec th

5. kausarsalley

|dw:1362873411456:dw| |dw:1362873506755:dw|

6. dangbeau

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?

7. inkyvoyd

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.

8. kausarsalley

i also think so!

9. dangbeau

hm... i have another answer $\sec ^{2} \theta=\sec \theta ^{2}$ it's equal, right?

10. dangbeau

i mean question

11. inkyvoyd

no, of course not. sec^2(th)= sec th*sec th sec(th^2)=sec(th*th)

12. NoelGreco

No, it's not correct.Re-write sec^2 x as (sec x)^2, then power and chain rules.

13. dangbeau

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

14. inkyvoyd
15. dangbeau

16. inkyvoyd

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)

17. inkyvoyd

does that make sense?

18. dangbeau

ah ok i got it

19. inkyvoyd

okay - so the chain rule makes more sense to you now?

20. dangbeau

yup, thanks sooo much ^^

21. inkyvoyd

You're welcome :)