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

Answer 1

Emily Ammerman

#(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)#

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Answer 2

Isaiah Allen

To use implicit differentiation to find ( \frac{dy}{dx} ) given ( 4x^2 - 2xy + 3y^2 = 8 ), follow these steps:

Differentiate both sides of the equation with respect to ( x ).
Treat ( y ) as a function of ( x ) and use the chain rule when differentiating terms involving ( y ).
Solve the resulting equation for ( \frac{dy}{dx} ).

Starting with ( 4x^2 - 2xy + 3y^2 = 8 ):

[ \frac{d}{dx}(4x^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(3y^2) = \frac{d}{dx}(8) ]

[ 8x - 2x\frac{dy}{dx} - 2y + 6y\frac{dy}{dx} = 0 ]

[ (8x - 2y) + (-2x + 6y)\frac{dy}{dx} = 0 ]

[ \frac{dy}{dx} = \frac{2y - 8x}{6y - 2x} ]

That's the derivative ( \frac{dy}{dx} ) expressed in terms of ( x ) and ( y ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.