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    Ok, i have maths homework in for tomorrow, can anyone do differentiation? (obviously i haven't quite perfected technique yet);

    i need to find dy/dx when y= (116-21x^2) divided by 120.

    Anyclues?
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    (Original post by CoraBelle)
    Ok, i have maths homework in for tomorrow, can anyone do differentiation? (obviously i haven't quite perfected technique yet);

    i need to find dy/dx when y= (116-21x^2) divided by 120.

    Anyclues?
    i'm pretty sure it's just -42x/120 = -7x/20
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    (Original post by boygenious)
    i'm pretty sure it's just -42x/120 = -7x/20
    Thanks so much, yup thats what is said.. i see it now they just cancelled down...

    I don't suppose you would be able to explain how you would find the point on a curve which is parrellel to a line?

    How woud you know if a line was a tangent to a curve?

    Sorry for all these questions - our teachers going mega fast through this and i'm more of a slow learner myself!

    thanks anyhow!
    Love CoraBelle
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    (Original post by CoraBelle)
    Thanks so much, yup thats what is said.. i see it now they just cancelled down...

    I don't suppose you would be able to explain how you would find the point on a curve which is parrellel to a line?

    How woud you know if a line was a tangent to a curve?

    Sorry for all these questions - our teachers going mega fast through this and i'm more of a slow learner myself!

    thanks anyhow!
    Love CoraBelle
    Hey

    When the curve is parallel to the line then the gradients are equal, so dy/dx are equal and then you can work out x.

    similar thing for if a line is a tangent to the curve. the tangent and the line have the same gradient so dy/dx is equal!

    Sorry that was a bit brief. Hope it helps
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    (Original post by CoraBelle)
    Ok, i have maths homework in for tomorrow, can anyone do differentiation? (obviously i haven't quite perfected technique yet);

    i need to find dy/dx when y= (116-21x^2) divided by 120.

    Anyclues?
    y = (116-21x^2)/120
    = 116/120 - (21/210)x^2
    dy/dx = -2(21/210)x
    = -5x

    whatever mik1a says, goes.
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    unfortunately not everything mik1a says goes... hes just plain wrong. that is not how you differentiate that function. this first reply was perfectly correct. look again mik1a and dazya
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    (Original post by Ally45)
    unfortunately not everything mik1a says goes... hes just plain wrong. that is not how you differentiate that function. this first reply was perfectly correct. look again mik1a and dazya
    what? please explain this to me sum1
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    I am perturbed by the fact that 2 different answers are given.
    Nothing upsets me more than not understanding maths. It's like the world has gone mad!
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    (Original post by mik1a)
    y = (116-21x^2)/120
    = 116/120 - (21/210)x^2
    dy/dx = -2(21/210)x
    = -5x
    Dude, she said the answer said something else, that isn't your answer.
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    (Original post by mik1a)
    y = (116-21x^2)/120
    = 116/120 - (21/210)x^2
    dy/dx = -2(21/210)x
    = -5x
    errm,no.Its y=116/120 - (21x^2)/120 => dy/dx=-42x/120
    were did (21/210)x^2 come from?
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    (Original post by IntegralAnomaly)
    errm,no.Its y=116/120 - (21x^2)/120 => dy/dx=-42x/120
    were did (21/210)x^2 come from?
    (21x^2)/120 = (21/120)(x^2*1) apparently...
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    (Original post by XTinaA)
    (21x^2)/120 = (21/120)(x^2*1) apparently...
    It does??!! wow u must shoe me the proof
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    (Original post by IntegralAnomaly)
    It does??!! wow u must shoe me the proof
    So you're being sarcastic, yeah?
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    Ah, the world makes sense again!
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    (Original post by XTinaA)
    So you're being sarcastic, yeah?
    erhmmm,sorry about that
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    (Original post by CoraBelle)
    Ok, i have maths homework in for tomorrow, can anyone do differentiation? (obviously i haven't quite perfected technique yet);

    i need to find dy/dx when y= (116-21x^2) divided by 120.

    Anyclues?
    isnt it just dy/dx=-42x/120=-7x/20????
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    (Original post by CoraBelle)
    Thanks so much, yup thats what is said.. i see it now they just cancelled down...

    I don't suppose you would be able to explain how you would find the point on a curve which is parrellel to a line?

    How woud you know if a line was a tangent to a curve?

    Sorry for all these questions - our teachers going mega fast through this and i'm more of a slow learner myself!

    thanks anyhow!
    Love CoraBelle
    If a point on a curve is parallel to a line then it follows that the gradient of the curve at that point is the same as the gradient of the line,if that just wooshed past ur head draw it and then u'll c
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    Quote:
    --------------------------------------------------------------------------------
    Originally Posted by Ally45
    unfortunately not everything mik1a says goes... hes just plain wrong. that is not how you differentiate that function. this first reply was perfectly correct. look again mik1a and dazya
    --------------------------------------------------------------------------------

    what? please explain this to me sum1

    all i was saying mate is that mik1a is wrong.... hes confused himself somewhere! if you dont understand why i've siad this, hit the books and go back to differentiation from first principles.
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    (Original post by Ally45)
    Quote:
    --------------------------------------------------------------------------------
    Originally Posted by Ally45
    unfortunately not everything mik1a says goes... hes just plain wrong. that is not how you differentiate that function. this first reply was perfectly correct. look again mik1a and dazya
    --------------------------------------------------------------------------------

    what? please explain this to me sum1
    dont worry...i wasnt saying that mik1a was right or nething...just confused with every1 seemingly putting different answers to wot was surely a normal differentiation q...i put my answer further up and it seems correct
    all i was saying mate is that mik1a is wrong.... hes confused himself somewhere! if you dont understand why i've siad this, hit the books and go back to differentiation from first principles.
 
 
 
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