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1. Determine the derivative using implicit differentiation:

The equation provided is: $2x - y^2 + 2y^3 = 1 + x^5$

To find $\frac{dy}{dx}$, we'll differentiate both sides with respect to $x$, treating $y$ as an implicit function of $x$.

Step-by-Step Differentiation:

1. Differentiate $2x$ with respect to $x$: $\frac{d}{dx}(2x) = 2$

2. Differentiate $-y^2$ with respect to $x$: $\frac{d}{dx}(-y^2) = -2y \cdot \frac{dy}{dx}$

3. Differentiate $2y^3$ with respect to $x$: $\frac{d}{dx}(2y^3) = 2 \cdot 3y^2 \cdot \frac{dy}{dx} = 6y^2 \cdot \frac{dy}{dx}$

4. Differentiate $1$ with respect to $x$: $\frac{d}{dx}(1) = 0$

5. Differentiate $x^5$ with respect to $x$: $\frac{d}{dx}(x^5) = 5x^4$

Putting it all together, we get: $2 - 2y \cdot \frac{dy}{dx} + 6y^2 \cdot \frac{dy}{dx} = 5x^4$

Combine like terms involving $\frac{dy}{dx}$: $2 + (6y^2 - 2y) \cdot \frac{dy}{dx} = 5x^4$

Now, solve for $\frac{dy}{dx}$: $(6y^2 - 2y) \cdot \frac{dy}{dx} = 5x^4 - 2$ $\frac{dy}{dx} = \frac{5x^4 - 2}{6y^2 - 2y}$

2. Find the equation of the tangent line to the curve at a specific point:

The curve given is: $y^5 + x^3 = y^2 + 12x$ and we are asked to find the equation of the tangent line at the point $(0, 1)$.

Step-by-Step Differentiation:

1. Differentiate $y^5$ with respect to $x$: $\frac{d}{dx}(y^5) = 5y^4 \cdot \frac{dy}{dx}$

2. Differentiate $x^3$ with respect to $x$: $\frac{d}{dx}(x^3) = 3x^2$

3. Differentiate $y^2$ with respect to $x$: $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$

4. Differentiate $12x$ with respect to $x$: $\frac{d}{dx}(12x) = 12$

Putting it all together, we get: $5y^4 \cdot \frac{dy}{dx} + 3x^2 = 2y \cdot \frac{dy}{dx} + 12$

Combine terms involving $\frac{dy}{dx}$: $(5y^4 - 2y) \cdot \frac{dy}{dx} = 12 - 3x^2$

Now, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{12 - 3x^2}{5y^4 - 2y}$

Calculate the Slope at $(0, 1)$:

Substitute $x = 0$ and $y = 1$: $\frac{dy}{dx} = \frac{12 - 3 \cdot 0^2}{5 \cdot 1^4 - 2 \cdot 1} = \frac{12}{5 - 2} = \frac{12}{3} = 4$

Equation of the Tangent Line:

Using the point-slope form $y - y_1 = m(x - x_1)$, where $m = 4$, $x_1 = 0$, and $y_1 = 1$: $y - 1 = 4(x - 0)$ $y = 4x + 1$

Summary of Answers:

1. $\frac{dy}{dx} = \frac{5x^4 - 2}{6y^2 - 2y}$
2. The equation of the tangent line at $(0, 1)$ is $y = 4x + 1$.

Would you like further details or explanations on any steps?

Here are five related questions that might deepen your understanding:

1. What is the geometric interpretation of implicit differentiation?
2. How can implicit differentiation be used for higher-order derivatives?
3. Why does finding the tangent line to a curve involve differentiating implicitly?
4. How would you find normal lines instead of tangent lines using these methods?
5. What are some real-world applications of implicit differentiation?

Tip: Always substitute the point into your derivative formula before solving for the tangent line equation to avoid algebraic mistakes.