4) Find the equation of the tangent line to the graph of $$ 8 y^{4}+8 x y^{24}+6 x^{3}-64=8 y^{2} $$ at $$ (2,-1) $$.

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To find the equation of the tangent line to the graph of

$$ 8 y^{4} + 8 x y^{24} + 6 x^{3} - 64 = 8 y^{2} $$

at the point $$(2, -1)$$, follow these steps:

### 1. Differentiate Implicitly

Differentiate both sides of the equation with respect to $$x$$:

$$
\frac{d}{dx}\left(8 y^{4}\right) + \frac{d}{dx}\left(8 x y^{24}\right) + \frac{d}{dx}\left(6x^{3}\right) - \frac{d}{dx}\left(64\right) = \frac{d}{dx}\left(8 y^{2}\right)
$$

Compute each derivative term-by-term:

- $$\frac{d}{dx}(8 y^{4}) = 32 y^{3} \frac{dy}{dx}$$
- $$\frac{d}{dx}(8 x y^{24}) = 8 y^{24} + 192 x y^{23} \frac{dy}{dx}$$ (using the product rule)
- $$\frac{d}{dx}(6x^{3}) = 18x^{2}$$
- $$\frac{d}{dx}(64) = 0$$
- $$\frac{d}{dx}(8 y^{2}) = 16 y \frac{dy}{dx}$$

Putting it all together:

$$
32 y^{3} \frac{dy}{dx} + 8 y^{24} + 192 x y^{23} \frac{dy}{dx} + 18x^{2} = 16 y \frac{dy}{dx}
$$

### 2. Solve for $$\frac{dy}{dx}$$

Collect all terms involving $$\frac{dy}{dx}$$ on one side:

$$
\left(32 y^{3} + 192 x y^{23} - 16 y\right) \frac{dy}{dx} = -8 y^{24} - 18x^{2}
$$

Thus,

$$
\frac{dy}{dx} = \frac{-8 y^{24} - 18x^{2}}{32 y^{3} + 192 x y^{23} - 16 y}
$$

### 3. Evaluate at $$(2, -1)$$

Substitute $$x = 2$$ and $$y = -1$$:

**Numerator:**

$$
-8 (-1)^{24} - 18(2)^{2} = -8(1) - 72 = -80
$$

**Denominator:**

$$
32 (-1)^{3} + 192(2)(-1)^{23} - 16(-1) = -32 - 384 + 16 = -400
$$

So,

$$
\frac{dy}{dx}\bigg|_{(2, -1)} = \frac{-80}{-400} = \frac{1}{5}
$$

### 4. Write the Tangent Line Equation

Using the point-slope form of a line:

$$
y - y_1 = m(x - x_1)
$$

Substitute $$m = \frac{1}{5}$$ and $$(x_1, y_1) = (2, -1)$$:

$$
y - (-1) = \frac{1}{5}(x - 2) \
y + 1 = \frac{1}{5}x - \frac{2}{5} \
y = \frac{1}{5}x - \frac{7}{5}
$$

**Final Equation:**

$$
y = \frac{1}{5}x - \frac{7}{5}
$$

**Alternatively**, multiplying through by 5 to eliminate fractions:

$$
x - 5y = 7
$$

**Answer:**

After simplifying, an equation for the tangent is y = (1⁄5)x – 7⁄5
The equation of the tangent line at the point (2, -1) is $$ y = \frac{1}{5}x - \frac{7}{5} $$.

Find the equation of the tangent line to the graph of at
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