To find $\frac{dy}{dx}$ by implicit differentiation of the equation $x^2 - 4xy + y^2 = 4$, follow these steps:

1. Differentiate both sides with respect to $x$, keeping in mind that $y$ is a function of $x$, so you must apply the chain rule when differentiating terms involving $y$.

$\frac{d}{dx}(x^2 - 4xy + y^2) = \frac{d}{dx}(4)$

2. Differentiate each term:

• $\frac{d}{dx}(x^2) = 2x$
• $\frac{d}{dx}(-4xy)$: Use the product rule, $\frac{d}{dx}(uv) = u'v + uv'$, where $u = -4x$ and $v = y$. $\frac{d}{dx}(-4xy) = (-4)(y) + (-4x)\left(\frac{dy}{dx}\right) = -4y - 4x\frac{dy}{dx}$
• $\frac{d}{dx}(y^2)$: Using the chain rule, $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$.

The right side of the equation is a constant, so $\frac{d}{dx}(4) = 0$.

Now we have: $2x - 4y - 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$

3. Solve for $\frac{dy}{dx}$:

Group the terms involving $\frac{dy}{dx}$ on one side: $- 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x + 4y$

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(-4x + 2y) = -2x + 4y$

Finally, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-2x + 4y}{-4x + 2y}$

Simplifying the expression: $\frac{dy}{dx} = \frac{2(y - x)}{2y - 4x} = \frac{y - x}{y - 2x}$

Thus, the derivative is: $\frac{dy}{dx} = \frac{y - x}{y - 2x}$

Would you like more details or further clarification on any part of this solution?

Here are five related questions to expand your understanding:

1. How would you apply implicit differentiation to more complex functions?
2. What happens when we differentiate trigonometric implicit functions?
3. How can you verify that the expression for $\frac{dy}{dx}$ is correct by substituting specific values for $x$ and $y$?
4. How does implicit differentiation handle higher-order derivatives?
5. Can implicit differentiation be used to find tangent lines to curves at specific points?

Tip: When performing implicit differentiation, remember to apply the chain rule carefully whenever differentiating a function involving $y$, since $y$ depends on $x$.