[mcfc2009]

Having trouble with 2. b) in getting the last mark.

http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN11.PDF

http://store.aqa.org.uk/qual/gce/pdf...W-MS-JUN11.PDF

Cheers.

[raheem94]

(Original post by mcfc2009)
Having trouble with 2. b) in getting the last mark.

http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN11.PDF

http://store.aqa.org.uk/qual/gce/pdf...W-MS-JUN11.PDF

Cheers.

Was this the confusion?

[mcfc2009]

(Original post by raheem94)


Was this the confusion?

Thanks, obvious!

[opiumJ]

Do you mean from:

6cos3x+6/(1+cos3x)^2

to

6/1+cos3x



Because if that's the last mark then 6cos3x+6 can be re-written as 6+6k (let's say cos3x is k).

so 6+6k/(1+k)^2 which is equal to:

6(1+k)/(1+k)(1+k)

take away a (1+k) from top and bottom and you are left with 6/1+k which is:

6/1+cos3x




Somehow, if you made it that far, I doubt that is the step you were stuck at so apologies if I just explained something basic.

[mcfc2009]

Can anyone help me with 7b
http://store.aqa.org.uk/qual/gceasa/...W-MS-JAN10.PDF
http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF

cheers

[GOT111]

(Original post by mcfc2009)
Can anyone help me with 7b
http://store.aqa.org.uk/qual/gceasa/...W-MS-JAN10.PDF
http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF

cheers

Try to rearrange the dy/dx into p(1+(tan4x)(tan4x))
This can then be used in the product rule (or chain, i can't remember which is which.)

Once you find that rearrange and remember y=tan4x