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    Hi, could someone please help with this differentiation C1 question?

    "Find the equation of the tangent to the curve at the given point:
    c) y=1/(3x^2) at (1/3, 3)

    I'm not sure where I'm going wrong!

    when I make y=(3x^2)^-1
    then so y = (1/3)x^-2
    so dy/dx =(-2/3)x^-3
    = -2/3x^3

    subbing x=1/3 into the derivative I get
    dy/dx = -2/1
    = -2 ????????????

    y - 3 = -2(x-1/3)
    y - 3 = -2x +2/3
    y = (11/3)- 2x

    or 3y + 2x = 11 ?

    But the answer in the book is y + 18x = 9 ?!?

    It's from OCR (non MEI) exc 8A 3c (c1)

    Attached the question as well

    Thanks in advanced
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    (Original post by Jitesh)
    Hi, could someone please help with this differentiation C1 question?

    "Find the equation of the tangent to the curve at the given point:
    c) y=1/(3x^2) at (1/3, 3)

    I'm not sure where I'm going wrong!

    when I make y=(3x^2)^-1
    then so y = (1/3)x^-2
    so dy/dx =(-2/3)x^-3
    = -2/3x^-3

    subbing x=1/3 into the derivative I get
    dy/dx = -2/1
    = -2 ????????????

    y - 3 = -2(x-1/3)
    y - 3 = -2x +2/3
    y = (11/3)- 2x

    or 3y + 2x = 11 ?

    But the answer in the book is y + 18x = 9 ?!?

    It's from OCR (non MEI) exc 8A 3c (c1)

    Attached the question as well

    Thanks in advanced
    It's where you've subbed x into dy/dx - the expression for dy/dx is correct but what you got out isn't.
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    (Original post by SeanFM)
    It's where you've subbed x into dy/dx - the expression for dy/dx is correct but what you got out isn't.
    Was it the -2/3x^-2 bit? I realised I think I typed that wrong as it should have been -2x^-2/3 or -2/3x^2

    I think I did that in my working out, I'll double check now
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    Wait I see where I went wrong, redid the calculation and got the gradient as -6, but wouldn't the gradient be -18 so is my dy/dx expression wrong?
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    (Original post by Jitesh)
    Was it the -2/3x^-2 bit? I realised I think I typed that wrong as it should have been -2x^-2/3 or -2/3x^2

    I think I did that in my working out, I'll double check now
    Not at all. dy/dx is exactly what you've written. It's just that you've put in x = 1/3, which is also correct, but got the wrong number out.

    It may help to break it down and show your steps as to how you put in x = 1/3.
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    (Original post by Jitesh)
    Wait I see where I went wrong, redid the calculation and got the gradient as -6, but wouldn't the gradient be -18 so is my dy/dx expression wrong?
    Your expression isn't wrong - please show how you put in x = 1/3 and we'll see what's going on, as you should indeed be getting -18 out and the only thing that's going wrong is the actual evaluation.
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    (Original post by SeanFM)
    Not at all. dy/dx is exactly what you've written. It's just that you've put in x = 1/3, which is also correct, but got the wrong number out.

    It may help to break it down and show your steps as to how you put in x = 1/3.
    Oh thanks SeanName:  12166423_10206762118584646_424537295_n.jpg
Views: 70
Size:  33.8 KB
    I tried skipping steps and did x^2 which was giving me 1/9 in my head

    Thanks again
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    (Original post by Jitesh)
    Oh thanks SeanName:  12166423_10206762118584646_424537295_n.jpg
Views: 70
Size:  33.8 KB
    I tried skipping steps and did x^2 which was giving me 1/9 in my head

    Thanks again
    Perfect! Well done Jitesh :borat:
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    (Original post by SeanFM)
    Perfect! Well done Jitesh :borat:
    Thank you! Sorry about that rookie error!
 
 
 
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