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Differentiation with negative integer indices

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?
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 :smile:
Reply 2
Avatar for marmbite
marmbite
OP
9
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
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.
(edited 8 years ago)
Reply 4
Avatar for marmbite
marmbite
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9
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
Reply 5
Avatar for razzor
razzor
10
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
13x2=13x−2 \frac{1}{3x^{2}} = \frac{1}{3}x^{-2}

Now differentiate as normal.
Reply 6
Avatar for marmbite
marmbite
OP
9
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
13x2=13x−2 \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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