To find the equation of the tangent line at $(1, 3)$ for the given curve $4x^2 - xy = 1$, follow these steps:

Step 1: Differentiate the equation implicitly

The equation is: $4x^2 - xy = 1$

Take the derivative of both sides with respect to $x$: $\frac{d}{dx}(4x^2) - \frac{d}{dx}(xy) = \frac{d}{dx}(1)$

Using the product rule for $-xy$ and the derivative rules: $8x - (x\frac{dy}{dx} + y) = 0$

Simplify: $8x - x\frac{dy}{dx} - y = 0$

Rearrange for $\frac{dy}{dx}$: $x\frac{dy}{dx} = 8x - y$

$\frac{dy}{dx} = \frac{8x - y}{x}$

Step 2: Evaluate the derivative at $(1, 3)$

Substitute $x = 1$ and $y = 3$ into the derivative formula: $\frac{dy}{dx} = \frac{8(1) - 3}{1} = \frac{8 - 3}{1} = 5$

The slope of the tangent line at $(1, 3)$ is $m = 5$.

Step 3: Write the equation of the tangent line

Using the point-slope form: $y - y_1 = m(x - x_1)$

Substitute $m = 5$, $x_1 = 1$, and $y_1 = 3$: $y - 3 = 5(x - 1)$

Final Answer:

The equation of the tangent line is: $\boxed{D. \, y - 3 = 5(x - 1)}$

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Would you like further explanation? Here are 5 related questions to deepen your understanding:

1. How does implicit differentiation differ from regular differentiation?
2. Why is the product rule necessary for differentiating $xy$?
3. What is the general form of the equation of a line, and how does it relate to point-slope form?
4. Can the tangent line at a point ever fail to exist for certain types of curves?
5. What happens if the given point does not lie on the curve? How would you verify it?

Tip: Always verify that the point lies on the curve by substituting its coordinates into the equation before proceeding with further calculations.