# How do you use implicit differentiation to find #(dy)/(dx)# given #5y^2=2x^3-5y#?

using the Chain Rule,

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To use implicit differentiation to find ( \frac{dy}{dx} ) given ( 5y^2 = 2x^3 - 5y ), follow these steps:

- Differentiate both sides of the equation with respect to ( x ).
- Use the chain rule where necessary.
- Solve the resulting equation for ( \frac{dy}{dx} ).

Differentiating both sides with respect to ( x ) gives:

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

Using the chain rule, differentiate each term:

[ \frac{d}{dx}(5y^2) = 10y \cdot \frac{dy}{dx} ]

[ \frac{d}{dx}(2x^3 - 5y) = 6x^2 - 5 \frac{dy}{dx} ]

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

[ 10y \cdot \frac{dy}{dx} = 6x^2 - 5 \frac{dy}{dx} ]

[ 10y \cdot \frac{dy}{dx} + 5 \frac{dy}{dx} = 6x^2 ]

[ (10y + 5) \frac{dy}{dx} = 6x^2 ]

[ \frac{dy}{dx} = \frac{6x^2}{10y + 5} ]

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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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