# How do you use implicit differentiation to find #(dy)/(dx)# given #sin2x^2y^3=3x^3+1#?

# dy/dx = (9x^2 - 4xcos(2x^2)y^3)/(3sin(2x^2)y^2) #

Example:

When this is done in situ it is known as implicit differentiation.

Now, we have:

Advanced Calculus

And so:

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To find ((dy)/(dx)) using implicit differentiation, follow these steps:

- Differentiate both sides of the equation with respect to (x).
- Apply the chain rule whenever you differentiate terms involving (y).
- Solve the resulting equation for ((dy)/(dx)).

Applying these steps to the given equation (sin(2x^2y^3) = 3x^3 + 1), we get:

[ \frac{d}{dx}(sin(2x^2y^3)) = \frac{d}{dx}(3x^3 + 1) ]

This simplifies to:

[ cos(2x^2y^3) \cdot \frac{d}{dx}(2x^2y^3) = 9x^2 ]

Then, using the chain rule:

[ cos(2x^2y^3) \cdot (4x^3y^2 \cdot \frac{dy}{dx} + 6x^2y^2) = 9x^2 ]

Solve for ((dy)/(dx)):

[ \frac{dy}{dx} = \frac{9x^2 - 6x^2y^2 \cdot cos(2x^2y^3)}{4x^3y^2 \cdot cos(2x^2y^3)} ]

This is the derivative of (y) 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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