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How do you implicitly differentiate # y^2= x^3 + 3x^2 #?

Answer 1

Aurora Alvarado

It depends on what you are using as the independent variable. For #t#, we get #2y dy/dt=3x^2 dx/dt+6x dx/dt#.

# y^2= x^3 + 3x^2 #

#d/dt( y^2) = d/dt(x^3 + 3x^2)#

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

Solve for the derivative you're looking for.

If the independent variable is taken to be #x#, then we get:

# y^2= x^3 + 3x^2 #

#d/dx( y^2) = d/dx(x^3 + 3x^2)#

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

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

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Answer 2

Adam Adkins

To implicitly differentiate the equation (y^2 = x^3 + 3x^2), follow these steps:

Differentiate both sides of the equation with respect to (x).
Use the chain rule when differentiating (y^2) with respect to (x).
Apply the power rule and chain rule when differentiating (x^3) and (3x^2), respectively.

The result of differentiating both sides with respect to (x) yields:

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

Solve for (\frac{dy}{dx}) to obtain the implicit derivative:

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

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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.