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To find the equation of the tangent line to the curve at \( x = 1 \), we need to follow these steps:

1. **Implicit Differentiation**: Differentiate the equation \( 3y^3 + xy - y = 81x^4 \) with respect to \( x \) to find \(\frac{dy}{dx}\).
   
   Given:
   \[
   3y^3 + xy - y = 81x^4
   \]

   Differentiating both sides with respect to \( x \):
   \[
   \frac{d}{dx}(3y^3) + \frac{d}{dx}(xy) - \frac{d}{dx}(y) = \frac{d}{dx}(81x^4)
   \]

   Using the chain rule and product rule:
   \[
   3 \cdot 3y^2 \cdot \frac{dy}{dx} + y + x \frac{dy}{dx} - \frac{dy}{dx} = 324x^3
   \]

   Simplify the terms to group \(\frac{dy}{dx}\):
   \[
   (9y^2 + x - 1)\frac{dy}{dx} = 324x^3 - y
   \]

   Solving for \(\frac{dy}{dx}\):
   \[
   \frac{dy}{dx} = \frac{324x^3 - y}{9y^2 + x - 1}
   \]

2. **Evaluate \( y \) when \( x = 1 \)**: Substitute \( x = 1 \) into the original equation to find the corresponding \( y \)-value.

   \[
   3y^3 + (1)(y) - y = 81(1)^4
   \]
   \[
   3y^3 = 81
   \]
   \[
   y^3 = 27
   \]
   \[
   y = 3
   \]

   So, when \( x = 1 \), \( y = 3 \).

3. **Calculate the Slope of the Tangent Line**: Substitute \( x = 1 \) and \( y = 3 \) into the derivative \(\frac{dy}{dx}\) to find the slope at that point.

   \[
   \frac{dy}{dx} = \frac{324(1)^3 - 3}{9(3)^2 + 1 - 1}
   \]
   \[
   = \frac{324 - 3}{81} = \frac{321}{81} = 4
   \]

   So, the slope of the tangent line at \( (1, 3) \) is 4.

4. **Equation of the Tangent Line**: Use the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), with \( (x_1, y_1) = (1, 3) \) and \( m = 4 \).

   \[
   y - 3 = 4(x - 1)
   \]
   \[
   y = 4x - 4 + 3
   \]
   \[
   y = 4x - 1
   \]

So, the equation of the tangent line is:
\[
y = 4x - 1
\]

Would you like further details on any step, or do you have additional questions?

Here are some related questions to explore:

1. How would you find the normal line to the curve at \( x = 1 \)?
2. What if \( x \) and \( y \) terms were higher powers? How would that affect differentiation?
3. How can you verify the point \( (1,3) \) lies on the original curve?
4. How does implicit differentiation differ from explicit differentiation in this problem?
5. Could we use a different method to find \(\frac{dy}{dx}\) here?

**Tip:** When using implicit differentiation, always remember to apply the product rule where necessary, especially when both \( x \) and \( y \) are in the same term.

Find the equation of the tangent line at the given value of x on the curve: 3y^3 + xy - y = 81x^4; x = 1

Step-by-step solution to finding the tangent line equation at x = 1 for the curve 3y^3 + xy - y = 81x^4. Learn implicit differentiation and apply the point-slope form to determine the tangent line.

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