You are Here: Home >< Maths

# Who can differentiate y = sec2x? watch

1. I think I know how to do it, but mine doesn't look like the answers

It's either:

a) -2(sec^2)2x
b) 2sec2xtan2x
c) sec2xtan2x
d) -2secxtan2x

Thanksssss
2. I reckon it's B, but my core 4 is rusty nowadays.
3. Core 3
4. I wanna say b) but someone more sure will probably clarify.
Sec(x) definately differentiates to sec(x)tan(x) and I assume chain rule still applies.
Core 3
Differentiating trig was core 4 for OCR
6. i'm pretty sure its b
you differentiate the X and put it at the front then keep the X with the sec and tan

2sec2xtan2x
7. using sub u=2x gives b...
8. (Original post by Bilkos_Ices)
Differentiating trig was core 4 for OCR
Ah I see, but I'm on AQA.
9. b

isnt secx in the formula book (well it is for aqa) so just 2x is going to leave you with b
10. Thanks guys
Thanks guys
By the way, you know 'd' is wrong because you've got a term in 'x' rather than '2x' (or it's a typo) so you should discard it immediately.
12. (Original post by jayshah31)
By the way, you know 'd' is wrong because you've got a term in 'x' rather than '2x' (or it's a typo) so you should discard it immediately.
It wasn't a typo. I thought it was b), I just needed a bit of clarification :]
13. I got (2tan2x)(sec2x)

which is equiv to B.

I can explain if you want?
14. (Original post by maxfire)
I got (2tan2x)(sec2x)

which is equiv to B.

I can explain if you want?
15. You can find the derivative by writing as . Using the quotient rule, differentiating gives you , which can be re-written as . HTH
16. sec(2x)

1/cos(2x)

[cos(2x)]^-1

-1[cos(2x)]^-2.-sin(2x).2

2[cos(2x)]^-2.sin(2x)

2sin(2x)/cos(2x)^2

2tan(2x).sex(2x)
I hope you realise that if you remove the bracket's it's exactly the same as b)

(Original post by nuodai)
You can find the derivative by writing as . Using the quotient rule, differentiating gives you , which can be re-written as . HTH

Less prone to mistakes
18. (Original post by nuodai)
You can find the derivative by writing as . Using the quotient rule, differentiating gives you , which can be re-written as . HTH
Don't judge me here, I've been self-teaching this module so I don't have a full understanding yet, but what's wrong with writing as (cos2x)^-1? Then using the chain rule to differentiate?

Edit: nevermind, many people have already posted the solution
19. (Original post by illy123)
sec(2x)

1/cos(2x)

[cos(2x)]^-1

-1[cos(2x)]^-2.-sin(2x).2

2[cos(2x)]^-2.sin(2x)

2sin(2x)/cos(2x)^2

2tan(2x).sex(2x)
Seems you drifted away a little on the last line there

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: July 19, 2009
Today on TSR

### He broke up with me because of long distance

Now I'm moving to his city

### University open days

1. Norwich University of the Arts
Thu, 19 Jul '18
2. University of Sunderland
Thu, 19 Jul '18
3. Plymouth College of Art
Thu, 19 Jul '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams