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# Differentiation with negative integer indices watch

1. 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?
2. 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)

Hope this helps
3. 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
4. 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.
5. (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
6. (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

Now differentiate as normal.
7. (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

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

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