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@FlyLikeFei: When I differentiate I always use negative exponents instead of fractions, then the regular rules still apply. Multiply the 1/3 by -3, and lower the exponent by a degree of 1. I just don't know why your answer says it's positive, it should be negative.

[quote=xVolcomStone]Nah, it's x^2.

Unless you meant (1/3)x^(-3), which would be -x^(-4)[/quote]
yeah that's what i meant, but how did you get it.

[quote=FlyLikeFei]the answer for the 1st is x^-4 but i dunno why that is[/quote]

Nah, it's x^2.

Unless you meant (1/3)x^(-3), which would be -x^(-4)

1/(3x^3)

x^(-3)/3
Derive
-3(x^-4)/3
x^(-4)

y=x^2+ -7/x^2

y'= 2x + -14x^(-3)
sub in x to get the slope of the tangent.

y=(slope of tangent)x + c

Use the x and y points to find C.

[quote=FlyLikeFei]the answer for the 1st is x^-4 but i dunno why that is[/quote]

Unless you initially wrote it wrong or I misunderstood what you wrote, that's wrong.
Using the power rule of differentiation, (x^3)/3 becomes x^2.

[quote=solarius]1) x^2
2) Differentiate and plug in the respective coordinates to find the slope. Then use the coordinates and the resulting slope in point slope form.[/quote]

the answer for the 1st is x^-4 but i dunno why that is

1) x^2
2) Differentiate and plug in the respective coordinates to find the slope. Then use the coordinates and the resulting slope in point slope form.

1/9x^2 is my guess. Although I might not have any idea as to what I'm talking about.

Don't you just plug in x,y and the gradient into (y2-y1) = m(x2-x1) ? Since it's at tangent the gradient of the curve has to equal to -1 when multiplied by the gradient of the tangent.