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

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Answer 1 · 2017-01-26T00:55:56

#(dy)/(dx) = (y-4x)/(3y-x)#

Differentiate both sides of the equation with respect to #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)#