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How do you use implicit differentiation to find #(dy)/(dx)# given #3x^2y+2xy^3=1#?

Answer 1

Charles Anctil

#dy/dx=(-2y^3-6xy)/(3x^2+6xy^2)#

differentiate with respect to x, using the #color(blue)"product rule"#

#3x^2dy/dx+6xy+2x.3y^2dy/dx+2y^3=0#

#rArr3x^2dy/dx+6xy+6xy^2dy/dx+2y^3=0#

#rArrdy/dx(3x^2+6xy^2)=-2y^3-6xy#

#rArrdy/dx=(-2y^3-6xy)/(3x^2+6xy^2)#

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Answer 2

Axel Alvarez

To find (\frac{dy}{dx}) using implicit differentiation for the equation (3x^2y + 2xy^3 = 1), 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 where necessary.
After differentiating, solve for (\frac{dy}{dx}).

Differentiating both sides with respect to (x), we get:

[6xy + 3x^2\frac{dy}{dx} + 2y^3 + 6xy^2\frac{dy}{dx} = 0]

Rearranging terms and solving for (\frac{dy}{dx}), we have:

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