The problem is asking to find $y_x'$ (the derivative of $y$ with respect to $x$) given the equation:

$x^4y^3 + \sin(x - 2y) + 5x^2 - y^2 = 0.$

We will differentiate this equation implicitly with respect to $x$:

1. Differentiate each term:

   • For $x^4y^3$, use the product rule: $\frac{d}{dx}(x^4y^3) = 4x^3y^3 + x^4 \cdot 3y^2 \frac{dy}{dx}.$
   • For $\sin(x - 2y)$, use the chain rule: $\frac{d}{dx}(\sin(x - 2y)) = \cos(x - 2y) \cdot (1 - 2\frac{dy}{dx}).$
   • For $5x^2$, the derivative is: $\frac{d}{dx}(5x^2) = 10x.$
   • For $-y^2$, use the chain rule: $\frac{d}{dx}(-y^2) = -2y \frac{dy}{dx}.$

2. Combine all terms: $4x^3y^3 + 3x^4y^2 \frac{dy}{dx} + \cos(x - 2y)(1 - 2\frac{dy}{dx}) + 10x - 2y \frac{dy}{dx} = 0.$

3. Collect terms involving $\frac{dy}{dx}$ on one side: $3x^4y^2 \frac{dy}{dx} - 2\cos(x - 2y)\frac{dy}{dx} - 2y\frac{dy}{dx} = -4x^3y^3 - \cos(x - 2y) - 10x.$

4. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx} \left( 3x^4y^2 - 2\cos(x - 2y) - 2y \right) = -4x^3y^3 - \cos(x - 2y) - 10x.$

5. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-4x^3y^3 - \cos(x - 2y) - 10x}{3x^4y^2 - 2\cos(x - 2y) - 2y}.$

This is the expression for $y_x'$.

Would you like a detailed explanation for any step or simplification?

Follow-up questions:

1. What are the principles behind implicit differentiation?
2. How does the product rule apply to composite functions?
3. Can you solve for $y_x'$ numerically for specific $x$ and $y$ values?
4. What is the significance of each term in the numerator and denominator of $y_x'$?
5. How would this approach change if the equation included logarithmic terms?

Tip: When solving implicit differentiation problems, always apply the chain rule carefully to terms involving $y$.