We are given the equation:

$\sqrt{x + y} = x^4 + y^4$

We need to find $\frac{dy}{dx}$ using implicit differentiation.

Step 1: Differentiate both sides with respect to $x$

• The left side contains a square root term, so we can rewrite it as $(x + y)^{1/2}$ and apply the chain rule.
• The right side contains $x^4 + y^4$, so we'll differentiate both terms separately.

Left-hand side:

$\frac{d}{dx}\left( (x + y)^{1/2} \right)$

Using the chain rule:

$\frac{1}{2}(x + y)^{-1/2} \cdot \left( \frac{d}{dx}(x + y) \right)$

Now, differentiate $x + y$ with respect to $x$:

$\frac{d}{dx}(x + y) = 1 + \frac{dy}{dx}$

So, the left-hand side becomes:

$\frac{1}{2}(x + y)^{-1/2} \cdot (1 + \frac{dy}{dx})$

Right-hand side:

$\frac{d}{dx}(x^4 + y^4)$

Differentiate both terms:

$\frac{d}{dx}(x^4) = 4x^3$ $\frac{d}{dx}(y^4) = 4y^3 \cdot \frac{dy}{dx}$

Thus, the right-hand side becomes:

$4x^3 + 4y^3 \cdot \frac{dy}{dx}$

Step 2: Set up the full equation

Now we combine the derivatives:

$\frac{1}{2}(x + y)^{-1/2} \cdot (1 + \frac{dy}{dx}) = 4x^3 + 4y^3 \cdot \frac{dy}{dx}$

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

We now need to solve this equation for $\frac{dy}{dx}$.

First, distribute the term on the left-hand side:

$\frac{1}{2}(x + y)^{-1/2} + \frac{1}{2}(x + y)^{-1/2} \cdot \frac{dy}{dx} = 4x^3 + 4y^3 \cdot \frac{dy}{dx}$

Now, move all terms involving $\frac{dy}{dx}$ to one side:

$\frac{1}{2}(x + y)^{-1/2} \cdot \frac{dy}{dx} - 4y^3 \cdot \frac{dy}{dx} = 4x^3 - \frac{1}{2}(x + y)^{-1/2}$

Factor out $\frac{dy}{dx}$ on the left-hand side:

$\frac{dy}{dx} \left( \frac{1}{2}(x + y)^{-1/2} - 4y^3 \right) = 4x^3 - \frac{1}{2}(x + y)^{-1/2}$

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

$\frac{dy}{dx} = \frac{4x^3 - \frac{1}{2}(x + y)^{-1/2}}{\frac{1}{2}(x + y)^{-1/2} - 4y^3}$

This is the derivative $\frac{dy}{dx}$ of the given equation.

Would you like more details or further clarifications?

Here are some related questions:

1. What does implicit differentiation mean, and when is it used?
2. How do you apply the chain rule in implicit differentiation?
3. Why do we need to move terms involving $\frac{dy}{dx}$ to one side?
4. What happens if $y$ is a constant with respect to $x$?
5. How can you interpret the result of $\frac{dy}{dx}$ in this context?

Tip: When working with implicit differentiation, remember to always treat $y$ as a function of $x$, even if it isn't explicitly solved for.