Scoobynet Maths bods.....
01 June 2005, 06:52 PM
#1
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Scoobynet Maths bods.....
Can anybody differentiate these for me please...
1. y= 6/x^2 - 7/x^3 + 12x^3/4
2. y= (x^2 -6x - 2) 1nx
3. y= (x^4 - 8x^2 - 4) / (x^2 + 5x + 9)
4. y= (x^2 + 6x - 3)^8
...because im stuck!
Much obliged as ever.
1. y= 6/x^2 - 7/x^3 + 12x^3/4
2. y= (x^2 -6x - 2) 1nx
3. y= (x^4 - 8x^2 - 4) / (x^2 + 5x + 9)
4. y= (x^2 + 6x - 3)^8
...because im stuck!
Much obliged as ever.
01 June 2005, 07:15 PM
#4
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Originally Posted by paul-s
christ 
way over my head. looks like japanese to me 
way over my head. looks like japanese to me Unfortunately i need the answer pretty sharpish otherwise im in the sh*t!
Anybody??
01 June 2005, 07:24 PM
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These don't look that hard? Although your notation is a bit confusing in places, so I might be misunderstanding the questions!
Number 1, I assume you mean 12x^(3/4)
Number 2, not sure what the 1nx means?
Number 1, I assume you mean 12x^(3/4)
Number 2, not sure what the 1nx means?
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01 June 2005, 07:28 PM
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Clue for number 1: apply the power rule, remember that 1/x^3 is the same as x^-3
Clue for number 3: apply the quotient rule
Number 4 could just be multiplied out.
Clue for number 3: apply the quotient rule
Number 4 could just be multiplied out.
01 June 2005, 07:30 PM
#9
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Originally Posted by Sprint Chief
These don't look that hard? Although your notation is a bit confusing in places, so I might be misunderstanding the questions!
Number 1, I assume you mean 12x^(3/4)
Number 2, not sure what the 1nx means?
Number 1, I assume you mean 12x^(3/4)
Number 2, not sure what the 1nx means?
No.2. Honestly dont have a clue what it means either. Maybe it just goes as a constant, but i really dont know.
I just cant remember how to do them but i sure when i see the answers i will be able to work backwards and get my head around them.
Cheers.
01 June 2005, 07:35 PM
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Number 4 could be done by substitution and application of the chain rule, which would be quicker than multiplying out, but you could always just let Comper100 work it out for you 
Not quite sure about Number 2 - if it was just times 1nx then why bother putting the 1 there? Confused!
Not quite sure about Number 2 - if it was just times 1nx then why bother putting the 1 there? Confused!
01 June 2005, 07:44 PM
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01 June 2005, 07:57 PM
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Oh and 3 with quoitent rule applied is: Hmm how am i going to write this?
dy/dx = ((x^2 - 5x - 9)(4x^3 - 16x) - (x^4 - 8x^2 - 4)(2x -5))/((x^2 - 5x - 9)^2))
Hows that
dy/dx = ((x^2 - 5x - 9)(4x^3 - 16x) - (x^4 - 8x^2 - 4)(2x -5))/((x^2 - 5x - 9)^2))
Hows that
01 June 2005, 08:02 PM
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Wouldn't the first part of 3 be
(4x^3 - 16x)/(x^2 + 5x + 9)
and the denonimator of the 2nd part
(x^2 + 5x + 9)^2
ie
(4x^3 - 16x)/(x^2 + 5x + 9) - ((x^4 - 8x^2 - 4)(2x -5))/((x^2 + 5x + 9)^2)
?
(4x^3 - 16x)/(x^2 + 5x + 9)
and the denonimator of the 2nd part
(x^2 + 5x + 9)^2
ie
(4x^3 - 16x)/(x^2 + 5x + 9) - ((x^4 - 8x^2 - 4)(2x -5))/((x^2 + 5x + 9)^2)
?
Last edited by krazy; 01 June 2005 at 08:14 PM.