The Student Room Group

Question 7a Jan 2010 C3

can someone please tell me if my working will gain full marks for question 7a from jan 2010 paper
the mark scheme has done it another way so i wasn't quite sure.

This is what i did:
secx=1/cosxsecx=1/cosx
u=1,v=cosxu=1, v=cosx
du/dx=0,dv/dx=sinxdu/dx=0, dv/dx=-sinx
using the quotient rule~ cosx(0)sinxcosx2 \frac{cosx(0)--sinx}{cosx^2}
=sinx/cosx2= sinx/cosx^2
=1/cosxsinx/cosx=secxtanx=1/cosx* sinx/cosx = secxtanx as required.

Thanks.
(edited 11 years ago)
Reply 1
Original post by gaffer dean
can someone please tell me if my working will gain full marks for question 7a from jan 2010 paper
the mark scheme has done it another way so i wasn't quite sure.

This is what i did:
secx=1/cosxsecx=1/cosx
u=1,v=cosxu=1, v=cosx
du/dx=0,dv/dx=sinxdu/dx=0, dv/dx=-sinx
using the quotient rule~ cosx(0)sinxcosx2 \frac{cosx(0)--sinx}{cosx^2}
=sinx/cosx2= sinx/cosx^2
=1/cosxsinx/cosx=secxtanx=1/cosx* sinx/cosx = secxtanx as required.

Thanks.


Yes that's correct as you've answered the question using the method described.

One thing.

You should write

ddxsecx=....\frac{d}{dx} \sec x =....

after saying 'using the quotient rule'. If it was y= you would say dy/dx=...

So you should do that when using any "rule".
(edited 11 years ago)
Reply 2
Original post by Dr A
Yes that's correct as you've answered the question using the method described.

One thing.

You should write

ddxsecx=....\frac{d}{dx} \sec x =....

after saying 'using the quotient rule'. If it was y= you would say dy/dx=...

So you should do that when using any "rule".

cool, yep will do,
cheers mate.
(edited 11 years ago)

Quick Reply

Latest