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How do you find dy/dx by implicit differentiation for #2x^3 + x^2*y - xy^3 = 6#?

Answer 1

Christopher Allers

You must remember that #y# is a function of #x# and derive it as well:

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Answer 2

Samantha Adkison

To find ( \frac{dy}{dx} ) by implicit differentiation for the equation ( 2x^3 + x^2y - xy^3 = 6 ):

Differentiate both sides of the equation with respect to ( x ).
Apply the chain rule whenever differentiating terms containing ( y ).
Isolate ( \frac{dy}{dx} ) on one side of the equation.

Here are the steps:

Differentiate each term with respect to ( x ):

( \frac{d}{dx}(2x^3) + \frac{d}{dx}(x^2y) - \frac{d}{dx}(xy^3) = \frac{d}{dx}(6) )
Apply the chain rule to terms involving ( y ):

( 6x^2 + 2xy + x^2\frac{dy}{dx} - y^3 - 3xy^2\frac{dy}{dx} = 0 )
Group terms involving ( \frac{dy}{dx} ) on one side:

( (2xy - 3xy^2 + x^2)\frac{dy}{dx} = y^3 - 6x^2 - 2xy )
Finally, solve for ( \frac{dy}{dx} ):

( \frac{dy}{dx} = \frac{y^3 - 6x^2 - 2xy}{2xy - 3xy^2 + x^2} )

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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.