# How do you differentiate #5xy + y^3 = 2x + 3y#?

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To differentiate (5xy + y^3 = 2x + 3y), you can use implicit differentiation. Differentiate each term with respect to (x) using the chain rule for terms containing (y).

Differentiate (5xy) with respect to (x):

[ \frac{d}{dx}(5xy) = 5y + 5x \frac{dy}{dx} ]

Differentiate (y^3) with respect to (x):

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

Differentiate (2x) and (3y) with respect to (x):

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

Then, equate the sum of these derivatives to zero since the equation is implicitly defined:

[ 5y + 5x \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 2 + 3 \frac{dy}{dx} ]

Now, isolate (\frac{dy}{dx}):

[ 5x \frac{dy}{dx} + 3y^2 \frac{dy}{dx} - 3 \frac{dy}{dx} = 2 - 5y ]

[ (5x + 3y^2 - 3) \frac{dy}{dx} = 2 - 5y ]

[ \frac{dy}{dx} = \frac{2 - 5y}{5x + 3y^2 - 3} ]

So, the derivative of the given equation is ( \frac{2 - 5y}{5x + 3y^2 - 3} ).

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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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