gaffer dean
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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/cosx
u=1, v=cosx
du/dx=0, dv/dx=-sinx
using the quotient rule~  \frac{cosx(0)--sinx}{cosx^2}
= sinx/cosx^2
=1/cosx* sinx/cosx = secxtanx as required.

Thanks.
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Dr A
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(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/cosx
u=1, v=cosx
du/dx=0, dv/dx=-sinx
using the quotient rule~  \frac{cosx(0)--sinx}{cosx^2}
= sinx/cosx^2
=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

\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".
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gaffer dean
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Report Thread starter 6 years ago
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(Original post by Dr A)
Yes that's correct as you've answered the question using the method described.

One thing.

You should write

\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.
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