How do you find dy/dx by implicit differentiation for #2x^3 + x^2*y  xy^3 = 6#?
You must remember that
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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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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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