We are given the equation

$\sqrt{x^2 + 4xy + y^6} = 2$

and are asked to find the derivative $\frac{dy}{dx}$, assuming that the equation implicitly defines $y$ as a differentiable function of $x$.

Step 1: Differentiate both sides implicitly

First, square both sides of the equation to eliminate the square root:

$x^2 + 4xy + y^6 = 4$

Now, differentiate both sides with respect to $x$.

Left-hand side:

We'll need to apply the product rule to the term $4xy$, and the chain rule for the term $y^6$, which involves $y$, a function of $x$.

1. $\frac{d}{dx}(x^2) = 2x$
2. $\frac{d}{dx}(4xy) = 4 \left( \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y) \right) = 4(y + x \frac{dy}{dx})$
3. $\frac{d}{dx}(y^6) = 6y^5 \cdot \frac{dy}{dx}$

So, the derivative of the left-hand side is:

$2x + 4(y + x \frac{dy}{dx}) + 6y^5 \frac{dy}{dx}$

Right-hand side:

The derivative of the constant $4$ is $0$, so the right-hand side simply becomes $0$.

Step 2: Set up the equation

Now, we have the equation:

$2x + 4(y + x \frac{dy}{dx}) + 6y^5 \frac{dy}{dx} = 0$

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

First, expand and group terms involving $\frac{dy}{dx}$:

$2x + 4y + 4x \frac{dy}{dx} + 6y^5 \frac{dy}{dx} = 0$

Now, isolate the terms involving $\frac{dy}{dx}$:

$4x \frac{dy}{dx} + 6y^5 \frac{dy}{dx} = -2x - 4y$

Factor out $\frac{dy}{dx}$:

$\left( 4x + 6y^5 \right) \frac{dy}{dx} = -2x - 4y$

Finally, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-2x - 4y}{4x + 6y^5}$

Thus, the derivative is:

$\frac{dy}{dx} = \frac{-2(x + 2y)}{4x + 6y^5}$

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Would you like a further breakdown of any step or clarification on a specific part? Also, here are a few related questions:

1. What is the derivative $\frac{dy}{dx}$ of a more complex implicit equation?
2. How does the chain rule apply to implicit differentiation?
3. How do you apply the product rule in implicit differentiation?
4. What is the difference between explicit and implicit differentiation?
5. How would you differentiate an equation that involves both $x$ and $y$ raised to powers?

Tip: When working with implicit differentiation, always remember to treat $y$ as a function of $x$ and apply the chain rule for any term involving $y$.