use implicit differentiation to find the slope of the tangent line to the curve of x^2/3+y^2/3=4 at the point (-1,3sqrt3)

Answer 1

taking the first

2/3 x dx + 2/3 ydy=0

dy/dx=-x/y at the point, 1/(3sqrt3)

Answer 2

((x^2)/3)+((y^2)/3)=4
d/dx 2x/3 + d/dx 2y/3 (y')= 0
y'= -(2x/3)-(2y/3)

at the point (-1,3sqrt3)

y'=(-2(-1)/3)-(2(3sqrt3)/3)=(2/3)-((6sqrt3)/3)=(2/3)-2sqrt3

Answer is (2/3)-2sqrt3 correct?

Answer 3

Is is x^(2/3) or is it (x^2)/3 ?
If it was the latter you could multiply both sides by 3 to get rid of any fractions straight away.

Brittany, y' = -(2x/3) / (2y/3) = -x/y
not y' = -(2x/3)-(2y/3)

Answer 4

it is x^(2/3)

Answer 5

The problem is (x^(2/3))+(y^(2/3))=4

I'm confused on the answer

Answer 6

I got the same answer as Brittany, I just want to make sure if it is correct or not

Answer 7

ok let's see...

x^(2/3) + y^(2/3) = 4
differentiating we get
(2/3)*(x^(-1/3)) + (2/3)*(y^(-1/3)).dy/dx = 0

divide across by (2/3) to get

x^(-1/3) + y^(-1/3).dy/dx = 0

Then replace the negative power with 1/ instead. e.g. x^-1 = 1/x

1/x^(1/3) + (1/y^(1/3)).dy/dx = 0
(1/y^(1/3)).dy/dx = -1/x^(1/3)

dy/dx = -y^(1/3)/x^(1/3) = -(y/x)^1/3

pt (1, 3sqrt3)

dy/dx = -(3sqrt3)^1/3

Bit nasty to be honest.

Answer 8

of course, you can simplify that last result to just √3, since it is (3^(3/2))^(1/3) = 3^(1/2)

Also, you dropped/gained a minus sign in there somewhere. The point is in QII, where the slope is positive.