Find the equation of the tangent line to the curve 2x6 + y4 = 9xy at the point (1, 2)

To find the equation of the tangent line to the curve \( 2x^6 + y^4 = 9xy \) at the point \( (1, 2) \), we need to follow these steps:

### Step 1: Differentiate implicitly with respect to \(x\).
We will take the derivative of both sides of the given equation implicitly, keeping in mind that \(y\) is a function of \(x\) (so we use the chain rule for terms involving \(y\)).

The equation is:
\[
2x^6 + y^4 = 9xy
\]

1. Differentiate \(2x^6\) with respect to \(x\):
   \[
   \frac{d}{dx}(2x^6) = 12x^5
   \]

2. Differentiate \(y^4\) with respect to \(x\):
   \[
   \frac{d}{dx}(y^4) = 4y^3 \frac{dy}{dx}
   \]
   (using the chain rule, since \(y\) is a function of \(x\)).

3. Differentiate \(9xy\) with respect to \(x\):
   \[
   \frac{d}{dx}(9xy) = 9 \left( y + x \frac{dy}{dx} \right)
   \]
   (using the product rule).

So, the implicit differentiation gives us the following equation:
\[
12x^5 + 4y^3 \frac{dy}{dx} = 9 \left( y + x \frac{dy}{dx} \right)
\]

### Step 2: Solve for \(\frac{dy}{dx}\).
Now, isolate \(\frac{dy}{dx}\) in the equation. First, distribute on the right-hand side:
\[
12x^5 + 4y^3 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx}
\]

Move all terms involving \(\frac{dy}{dx}\) to one side:
\[
4y^3 \frac{dy}{dx} - 9x \frac{dy}{dx} = 9y - 12x^5
\]

Factor out \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} (4y^3 - 9x) = 9y - 12x^5
\]

Solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{9y - 12x^5}{4y^3 - 9x}
\]

### Step 3: Evaluate \(\frac{dy}{dx}\) at the point \( (1, 2) \).
Substitute \( x = 1 \) and \( y = 2 \) into the derivative:
\[
\frac{dy}{dx} = \frac{9(2) - 12(1)^5}{4(2)^3 - 9(1)}
\]
\[
\frac{dy}{dx} = \frac{18 - 12}{32 - 9} = \frac{6}{23}
\]

So, the slope of the tangent line at the point \( (1, 2) \) is \( \frac{6}{23} \).

### Step 4: Find the equation of the tangent line.
Now that we know the slope of the tangent line is \( \frac{6}{23} \) and the line passes through \( (1, 2) \), we use the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( m = \frac{6}{23} \), \( x_1 = 1 \), and \( y_1 = 2 \):
\[
y - 2 = \frac{6}{23}(x - 1)
\]
Simplify:
\[
y = \frac{6}{23}(x - 1) + 2
\]
\[
y = \frac{6}{23}x - \frac{6}{23} + 2
\]
\[
y = \frac{6}{23}x + \frac{40}{23}
\]

Thus, the equation of the tangent line is:
\[
y = \frac{6}{23}x + \frac{40}{23}
\]

### Do you have any questions, or would you like a further breakdown of the steps?

Here are five related questions:
1. How do you apply implicit differentiation to other types of curves?
2. What are the applications of the tangent line in real-world problems?
3. How can you find the normal line to a curve at a given point?
4. What is the geometric meaning of the derivative at a point on a curve?
5. Can implicit differentiation be used for three-variable equations?

**Tip**: When applying implicit differentiation, always use the chain rule for terms involving dependent variables like \( y \).

Find the equation of the tangent line to the curve 2x^6 + y^4 = 9xy at the point (1, 2)

Learn how to find the equation of the tangent line to the curve 2x^6 + y^4 = 9xy at the point (1, 2). This solution covers implicit differentiation, derivatives, and the point-slope form. Ideal for high school students in Grades 11-12.

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