# What is the implicit derivative of #1= xy^3-3xy#?

It is

It is

#(1')=x'*y^3+x*3*y^2*y'-3x'y-3xy'=>
0=y^3+3xy^2*y'-3y-3xy'=>
y'(3x-3xy^2)=y^3-3y=>
dy/dx=(y^3-3y)/(3x-3xy^2)#

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To find the implicit derivative of the equation (1 = xy^3 - 3xy), differentiate both sides of the equation with respect to (x) using the product rule and chain rule where necessary:

[\frac{d}{dx} (1) = \frac{d}{dx} (xy^3 - 3xy)]

[0 = \frac{d}{dx} (xy^3) - \frac{d}{dx} (3xy)]

Using the product rule, we differentiate (xy^3) and (3xy) with respect to (x):

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

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

Now, substitute these derivatives back into the equation:

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

Rearrange the terms to solve for (\frac{dy}{dx}):

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

This is the implicit derivative of the equation (1 = xy^3 - 3xy) with respect to (x).

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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.

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.

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.

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