To find the equation of the tangent line to the curve given by [tex]\(x^6 y^6 = 64\)[/tex] at the point [tex]\((2, 1)\)[/tex], we will use implicit differentiation and then apply the point-slope form of the line equation. Here is the step-by-step process:
1. Given Equation:
[tex]\[ x^6 y^6 = 64 \][/tex]
2. Implicit Differentiation:
To find the derivative [tex]\(\frac{dy}{dx}\)[/tex], we differentiate both sides of the equation with respect to [tex]\(x\)[/tex], treating [tex]\(y\)[/tex] as a function of [tex]\(x\)[/tex] (i.e., [tex]\(y = y(x)\)[/tex]).
[tex]\[
\frac{d}{dx}(x^6 y^6) = \frac{d}{dx}(64)
\][/tex]
The right side is:
[tex]\[
\frac{d}{dx}(64) = 0
\][/tex]
Now, for the left side, we use the product rule:
[tex]\[
\frac{d}{dx}(x^6 y^6) = x^6 \frac{d}{dx}(y^6) + y^6 \frac{d}{dx}(x^6)
\][/tex]
Using the chain rule for [tex]\(y^6\)[/tex]:
[tex]\[
\frac{d}{dx}(y^6) = 6y^5 \frac{dy}{dx}
\][/tex]
And straightforward differentiation for [tex]\(x^6\)[/tex]:
[tex]\[
\frac{d}{dx}(x^6) = 6x^5
\][/tex]
So our equation becomes:
[tex]\[
x^6 \cdot 6y^5 \frac{dy}{dx} + y^6 \cdot 6x^5 = 0
\][/tex]
3. Solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
6x^6 y^5 \frac{dy}{dx} + 6x^5 y^6 = 0
\][/tex]
Factor out the common terms:
[tex]\[
6x^5 y^5 (x \frac{dy}{dx} + y) = 0
\][/tex]
Simplify and solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
x \frac{dy}{dx} + y = 0
\][/tex]
[tex]\[
\frac{dy}{dx} = -\frac{y}{x}
\][/tex]
4. Evaluate at the Point (2, 1):
We now find the slope of the tangent line at the point [tex]\((2, 1)\)[/tex]:
[tex]\[
\left. \frac{dy}{dx} \right|_{(2, 1)} = -\frac{y}{x} = -\frac{1}{2}
\][/tex]
5. Equation of the Tangent Line:
The point-slope form of the line equation is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Here, [tex]\((x_1, y_1) = (2, 1)\)[/tex] and the slope [tex]\(m = -\frac{1}{2}\)[/tex]:
[tex]\[
y - 1 = -\frac{1}{2}(x - 2)
\][/tex]
So, the equation of the tangent line to the curve [tex]\(x^6 y^6 = 64\)[/tex] at the point [tex]\((2, 1)\)[/tex] is:
[tex]\[
y - 1 = -\frac{1}{2}(x - 2)
\][/tex]
Or simplifying:
[tex]\[
y - 1 = -\frac{1}{2}x + 1
\][/tex]
[tex]\[
y = -\frac{1}{2}x + 2
\][/tex]
Therefore, the equation of the tangent line at the point [tex]\((2, 1)\)[/tex] is:
[tex]\[
y = -\frac{1}{2}x + 2
\][/tex]
