How do you use implicit differentiation to find dy/dx given $4 x^{2} - 2 x y + 3 y^{2} = 8$ $4x^2-2xy+3y^2=8$ ?

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Answer 1 · 2017-01-26T00:55:56

$\frac{d y}{d x} = \frac{y - 4 x}{3 y - x}$ $(dy)/(dx) = (y-4x)/(3y-x)$

Differentiate both sides of the equation with respect to $x$ $x$ , keeping in mind that:

$d/(dx) f(y(x)) = f'(y(x))* y'(x)$

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

$8x -2y -2xy' +6yy' = 0$

Solve now for $y'$ :

$2y'(3y-x)= 2y-8x$

$y' = (y-4x)/(3y-x)$

How do you use implicit differentiation to find dy/dx given 4x^2-2xy+3y^2=84x2−2xy+3y2=84x^2-2xy+3y^2=8?