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# Fp3

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1. http://mathsathawthorn.pbworks.com/f/FP3Jun06.pdf
q3a?
This is what ive done below!
f(x)= lnsecx
x=0 , f(x)=0

f'(x)=tanx
x=0, f'(x)=0

f''(x)=sec^2x
x=0, f''(x)=1

Is f'''(x)=2(sec^2xtanx)

And what is f''''(x)
2. You're right about f'''(x).

What have you tried to work out f''''(x)?
3. (Original post by Student403)
You're right about f'''(x).

What have you tried to work out f''''(x)?
Well can i use u=sec^2x and v= 2tanx? And find du/dx and dv/dx and cross multiply?
4. (Original post by Ayaz789)
Well can i use u=sec^2x and v= 2tanx? And find du/dx and dv/dx and cross multiply?
cross multiply? I don't see how you could cancel the du and dv. Could you explain how?
Spoiler:
Show
I'd use the quotient rule after putting tanx = sinx/cosx and secx = 1/cosx
5. (Original post by Student403)
You're right about f'''(x).

What have you tried to work out f''''(x)?
Or i can do u=2sec^2x and v=tanx as dv/dx =sec^2x but du/dx is?
6. (Original post by Student403)
cross multiply? I don't see how you could cancel the du and dv. Could you explain how?
Spoiler:
Show
I'd use the quotient rule after putting tanx = sinx/cosx and secx = 1/cosx
Like i do u* dv/dx and v* du/dx
7. (Original post by Ayaz789)
Or i can do u=2sec^x and v=tanx as dv/dx =sec^x but du/dx is?
I'm asking what would you do with du/dx and dv/dx? How do you cancel the du and dv? The substitution doesn't really make sense
8. (Original post by Student403)
I'm asking what would you do with du/dx and dv/dx? How do you cancel the du and dv? The substitution doesn't really make sense
I do u* dv/ dx???
And do v* du/ dx??
9. (Original post by Ayaz789)
I do u* dv/ dx???
And do v* du/ dx??
Oh you're talking about the product rule here. Yeah that's fine (sorry)

But remember since you have 2sec2x tanx and you want d/dx (2sec2x tanx), it's better to factor out the 2 to get

2 d/dx (sec2x tanx)
10. to find the fourth derivative it is sensible to replace sec2(x) with tan2(x) + 1

in the third derivative
11. (Original post by Student403)
Oh you're talking about the product rule here. Yeah that's fine

But remember since you have 2sec2x tanx and you want d/dx (2sec2x tanx), it's better to factor out the 2 to get

2 d/dx (sec2x tanx)
Okay can you show me how to get f''''(x)
12. (Original post by Ayaz789)
Okay can you show me how to get f''''(x)
No I can't show you, sorry.

Using the bear 's identity will work well
13. (Original post by Student403)
No I can't show you, sorry.

Using the bear 's identity will work well
Okay do me a favour and differentiate 2sec^2x
14. (Original post by Ayaz789)
Okay do me a favour and differentiate 2sec^2x
You can do that since you said

f''(x)=sec^2x

f'''(x)=2(sec^2xtanx)

So if you have 2sec2x, then what is that differentiated?
15. (Original post by Ayaz789)
Okay do me a favour and differentiate 2sec^2x
DIY
16. (Original post by Math12345)
DIY
& that means?
(Do it yourself)
17. Is f''''(X) = 2sec^4x +tanx(tanx/cosx)^2
18. (Original post by Ayaz789)
& that means?
(Do it yourself)
If you are really lazy, use wolframalpha to do it for you.
19. (Original post by Math12345)
If you are really lazy, use wolframalpha to do it for you.
Dw ive attempted it ^

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