We are given the equation:

$2 - 5y^3 - xy = x^2$

We need to find $\frac{dy}{dx}$ in terms of $x$ and $y$. To do this, we'll differentiate both sides of the equation implicitly with respect to $x$, treating $y$ as a function of $x$.

Step 1: Differentiate each term with respect to $x$

• For the constant $2$, its derivative is 0.

• For $-5y^3$, use the chain rule: $\frac{d}{dx}(-5y^3) = -5 \cdot 3y^2 \cdot \frac{dy}{dx} = -15y^2 \cdot \frac{dy}{dx}$

• For $-xy$, apply the product rule. The derivative of $xy$ is: $\frac{d}{dx}(-xy) = -\left(\frac{d}{dx}(x \cdot y)\right) = -(y + x \cdot \frac{dy}{dx})$

• For $x^2$, the derivative is: $\frac{d}{dx}(x^2) = 2x$

Step 2: Write the differentiated equation

Now, differentiating both sides of the equation:

$0 - 15y^2 \cdot \frac{dy}{dx} - (y + x \cdot \frac{dy}{dx}) = 2x$

Simplify this to:

$-15y^2 \cdot \frac{dy}{dx} - y - x \cdot \frac{dy}{dx} = 2x$

Step 3: Collect $\frac{dy}{dx}$ terms

Move the terms involving $\frac{dy}{dx}$ to one side of the equation:

$-15y^2 \cdot \frac{dy}{dx} - x \cdot \frac{dy}{dx} = 2x + y$

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

$\frac{dy}{dx}(-15y^2 - x) = 2x + y$

Step 4: Solve for $\frac{dy}{dx}$

Now, divide both sides by $(-15y^2 - x)$:

$\frac{dy}{dx} = \frac{2x + y}{-15y^2 - x}$

Thus, the derivative $\frac{dy}{dx}$ in terms of $x$ and $y$ is:

$\frac{dy}{dx} = \frac{2x + y}{-15y^2 - x}$

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Would you like more details on any of these steps or have any questions?

Here are 5 related questions:

1. How does implicit differentiation differ from explicit differentiation?
2. What is the product rule and how is it applied in calculus?
3. How can we interpret the physical meaning of $\frac{dy}{dx}$ in a real-world context?
4. Could we solve for $y$ explicitly from the original equation instead of using implicit differentiation?
5. What happens if we change the powers of $y$ or $x$ in the original equation?

Tip: When using the product rule in differentiation, remember that you must differentiate both the first and second functions in the product separately.