1. slope of tangent

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?

2. Your calculations are fine, haebinpark.

Maybe the problem was with the graphing.
The slopes of the tangents are 0.5 and -0.5,
which you will get if you simplify a little more

$\displaystyle \frac{\sqrt{2}}{\left(\frac{4}{\sqrt{2}}\right)}=\ frac{\sqrt{2}\sqrt{2}}{4}$

3. First find f(sqrt(2)) first which is sqrt(1/2)

x=sqrt(2)
y=sqrt(1/2)

derivative of x^2+4y^2=4

2x+8y*y'=0
2(x+4y*y')=0
x+4y*y'=0
y'=-x/(4y)