Questions: Find the slope of the tangent line to the curve (4 x^2+1 x y-4 y^3=-12) at the point ((2,2)).

Find the slope of the tangent line to the curve (4 x^2+1 x y-4 y^3=-12) at the point ((2,2)).
Transcript text: Find the slope of the tangent line to the curve $4 x^{2}+1 x y-4 y^{3}=-12$ at the point $(2,2)$.
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Solution

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Solution Steps

To find the slope of the tangent line to the curve \(4x^2 + xy - 4y^3 = -12\) at the point \((2, 2)\), we need to use implicit differentiation. Here are the high-level steps:

  1. Differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\) (i.e., use the chain rule for terms involving \(y\)).
  2. Solve for \(\frac{dy}{dx}\) to find the slope of the tangent line.
  3. Substitute the point \((2, 2)\) into the derivative to find the specific slope at that point.
Step 1: Implicit Differentiation

We start with the equation of the curve given by

\[ 4x^2 + xy - 4y^3 = -12. \]

Differentiating both sides with respect to \(x\) while applying the product rule and chain rule, we obtain:

\[ \frac{d}{dx}(4x^2) + \frac{d}{dx}(xy) - \frac{d}{dx}(4y^3) = 0. \]

This results in:

\[ 8x + y + x\frac{dy}{dx} - 12y^2\frac{dy}{dx} = 0. \]

Step 2: Solve for \(\frac{dy}{dx}\)

Rearranging the equation to isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx}(x - 12y^2) = -8x - y. \]

Thus, we can express \(\frac{dy}{dx}\) as:

\[ \frac{dy}{dx} = \frac{-8x - y}{x - 12y^2}. \]

Step 3: Evaluate at the Point \((2, 2)\)

Now, we substitute the point \((2, 2)\) into the derivative:

\[ \frac{dy}{dx} \bigg|_{(2, 2)} = \frac{-8(2) - 2}{2 - 12(2^2)} = \frac{-16 - 2}{2 - 48} = \frac{-18}{-46} = \frac{9}{23}. \]

Final Answer

The slope of the tangent line to the curve at the point \((2, 2)\) is

\[ \boxed{\frac{9}{23}}. \]

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