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    I can't get my head around this:

    Find the equation of the tangent to the curve at the given point.

    y=1/3x2 at (1/3, 3)
    Can somebody show me how this is done?
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    First, differentiate y=1/3x^2 (multiply 1/3 by 2 then decrease the power by 1).

    Then plug in the x-coordinate given, into dy/dx , to find m (the gradient)).

    Then put the values into y-y1 = m(x - x1)
    where x1=1/3 m=gradient and y1=3

    Hope this helps
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    Thanks.
    That was the method I tried, I got the gradient as -2 and the equation to 3y+6x=11
    But the answer is supposed to be y+18x=9
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    Is it supposed to be 1/3x^-2 rather than 1/3x^2?

    I tried it with the negative power and got dy/dx to equal -2/3(x)^3 and found the gradient to equal -18.

    I then subbed the values in again and came to y+18x=9.
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    (Original post by phoebelferg)
    Is it supposed to be 1/3x^-2 rather than 1/3x^2?

    I tried it with the negative power and got dy/dx to equal -2/3(x)^3 and found the gradient to equal -18.

    I then subbed the values in again and came to y+18x=9.

    Maybe that answer is for a negative power. The question definitely does not include a negative power, I've checked again. there is the possibility that the answer in the book is wrong.

    Thanks
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    (Original post by marmbite)
    Maybe that answer is for a negative power. The question definitely does not include a negative power, I've checked again. there is the possibility that the answer in the book is wrong.

    Thanks
    Sorry for the late response.
    In the question, is the x² part of the denominator? If so, then the answer in the book is correct; remember that
     \frac{1}{3x^{2}} = \frac{1}{3}x^{-2}

    Now differentiate as normal.
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    (Original post by razzor)
    Sorry for the late response.
    In the question, is the x² part of the denominator? If so, then the answer in the book is correct; remember that
     \frac{1}{3x^{2}} = \frac{1}{3}x^{-2}

    Now differentiate as normal.
    The x² is part of the denominator.
    Thanks for the help
 
 
 
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