Maths Question Help - GCSE!!!
23-04-2009, 10:45 AM
#46
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So 625x^2 times the power by the coefficient (625 x 2 = 1250) and then reduce the power by one to get 1250x
Same thing with the first term 5000x^1 (5000 x 1 = 5000) and reduce the power so x disappears.
I dont know how the hell this is on a gcse paper tho, I did it a few years ago and you only did this sort of thing at AS-level.
23-04-2009, 03:50 PM
#49
To differentiate it you times the power of x by the coefficient (the number in front of it) then reduce the power of x by 1.
So 625x^2 times the power by the coefficient (625 x 2 = 1250) and then reduce the power by one to get 1250x
Same thing with the first term 5000x^1 (5000 x 1 = 5000) and reduce the power so x disappears.
I dont know how the hell this is on a gcse paper tho, I did it a few years ago and you only did this sort of thing at AS-level.
So 625x^2 times the power by the coefficient (625 x 2 = 1250) and then reduce the power by one to get 1250x
Same thing with the first term 5000x^1 (5000 x 1 = 5000) and reduce the power so x disappears.
I dont know how the hell this is on a gcse paper tho, I did it a few years ago and you only did this sort of thing at AS-level.
So I was right then?
Rich
23-04-2009, 04:14 PM
#51
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That equation doesn't equal 0 its only the differentiated equation (5000 - 1250x) at the turning points that can equals 0, not the original one.
23-04-2009, 06:10 PM
#52
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well i did my GCSE's last year (got an A but not an A* due to my courseowrk being about an E) and that is not anything that we did/are doing this year in year 11
23-04-2009, 07:35 PM
#53
Too many posts.. I need a life!!
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Its from the following paper...
London Mathematics IGCSE
Mathematics
paper 3H
Higher Tier
Monday 6 November 2006 - morning
I was surprised as well.
She is back tonight for more revision so standby for a new question if they struggle
Thanks
Leigh
London Mathematics IGCSE
Mathematics
paper 3H
Higher Tier
Monday 6 November 2006 - morning
I was surprised as well.
She is back tonight for more revision so standby for a new question if they struggle
Thanks
Leigh
24-04-2009, 10:06 AM
#55
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Dividing both sides by x etc is wrong - the first answer is correct. Google 'differentiation'. General form is:
if y = ax^b
then dy/dx = (ab)x^(b-1), for each term
so if y= 1x + 1x^2 + 2x^3 + 3x^4
dy/dx = (1x1)x^(1-1) + (2x1)x^(2-1) + (3x2)x^(3-1) + (3x4)x^(4-1)
= 1x^0 + 2x^1 + 6x^2 + 12x^3
= 1 + 2x + 6x^2 + 12x^3
does that make sense?
if y = ax^b
then dy/dx = (ab)x^(b-1), for each term
so if y= 1x + 1x^2 + 2x^3 + 3x^4
dy/dx = (1x1)x^(1-1) + (2x1)x^(2-1) + (3x2)x^(3-1) + (3x4)x^(4-1)
= 1x^0 + 2x^1 + 6x^2 + 12x^3
= 1 + 2x + 6x^2 + 12x^3
does that make sense?
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