the question is

use implicit differentiation to find the slope of tangent at x = sqrt 2 for x^2 + 4y^2 = 4

what i did was

dx^2 / dx = 2x

d4y^2 / dx = 8yy'

d4/dx = 0

2x + 8yy' = 0

8yy' = -2x

y'= -2x/8y

= -x/4y

since x = sqrt 2

(sqrt 2)^2 + 4y^2 = 4

2+4y^2 = 4

4y^2 = 2

y^2 = 2/4 = 1/2

y= +/- sqrt 1/2

since y' = -x / 4y and x=sqrt 2, and y = +/- sqrt 1/2

y'= - sqrt2 / 4(sqrt 1/2)

and

y'= - sqrt2 / 4(- sqrt 1/2) = sqrt2 / 4(sqrt 1/2)

so i got slope of tangent = +/- sqrt2 / 4(sqrt 1/2)

but i drew graph on the calculator and seems like i did something wrong

but i can't find where i did go wrong...

does anyone see what i did wrong?