# (I) Find y' for y=y(x) defined implicitly by `3y^2-x^2+2x-y-3=0` . (II) And evaluate y' at (1,1).

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### 1 Answer

`3y^2-x^2+2x-y-3=0`

(I) To determine y', differentiate both sides of the equation with respect to x.

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

`6y dy/dx - 2x +2 - dy/dx - 0 = 0`

`6yy' - 2x + 2 -y' =0`

Then, group the terms with y'.

`(6yy' - y') - 2x + 2 = 0`

Factor the GCF of the group.

`y'(6y - 1) - 2x + 2 = 0`

Then, isolate y'.

`y'(6y - 1) -2x + 2x +2 - 2 = 0 + 2x - 2`

`y'(6y - 1) = 2x - 2`

`(y'(6y - 1))/(6y-1)= ( 2x - 2)/(6y-1)`

`y'=(2x-2)/(6y-1)`

**Hence, `y'=(2x-2)/(6y-1)` .**

(II) To solve for the value of y' at point (1,1), plug-in x=1 and y=1 to the derivative.

`y'=(2x-2)/(6y-1)`

`y'=(2*1-2)/(6*1-1)`

`y'=(2-2)/(6-1)`

`y'=0/5`

`y'=0`

**Hence, at (1,1), `y'=0` .**