Use implicit differentiation to find y'.  `a) (x+y)^2=4 ` `b) x^2y^2=16 ` `c)x^(1/2) + y^(1/2)=1` `` ``

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`a) (x+y)^2 = 4`

`==> x^2 + 2xy + y^2 = 4`

`==> (x^2)' + (2xy)' + (y^2)' = (4)'`

`==> 2x + (2xy'+2y) + 2yy' = 0`

`==> 2x + 2xy' + 2y + 2yy' = 0`

`==> 2xy'+2yy' = -2x-2y`

`==> y'(2x+2y)= -(2x+3y)`

`==> y' = -(2x+2y)/(2x+2y) = -1`

`==> y' =-1`

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`b) x^2 y^2 = 16`

`==> (x^2y^2)' = (16)'`

`==> (x^2)(y^2)' + (x^2)'(y^2) = 0`

`==> x^2 2yy' + 2x y^2 = 0`

`==> 2yx^2 y' = -2xy^2`

`==> y' = -(2xy^2)/(2yx^2)`

`==> y' = -y/x`

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`c) x^(1/2) + y^(1/2) = 1`

`==> (x^(1/2))' + (y^(1/2))' = 0`

`==> (1/2)x^(-1/2) + (1/2)y^(-1/2) y' = 0`

`==> 1/(2x^(1/2)) + (y')/(2y^(1/2)) = 0`

`==> (y')/(2y^(1/2)) = -1/(2x^(1/2))`

`==> y' = -(2y^(1/2))/(2x^(1/2))`

`==> y' = -y^(1/2)/x^(1/2) = - (y/x)^(1/2)`

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