To find $\frac{dy}{dx}$ using implicit differentiation for the equation

$4x^3y^2 - 10x^2y = 6,$

follow these steps:

1. Differentiate both sides with respect to $x$.

   • The derivative of $4x^3y^2$ requires the product rule: $\frac{d}{dx}(4x^3y^2) = 4(3x^2)y^2 + 4x^3(2y\frac{dy}{dx}) = 12x^2y^2 + 8x^3y\frac{dy}{dx}.$

   • The derivative of $-10x^2y$ also requires the product rule: $\frac{d}{dx}(-10x^2y) = -10(2x)y - 10x^2\frac{dy}{dx} = -20xy - 10x^2\frac{dy}{dx}.$

   • The derivative of $6$ is $0$.

2. Combine the derivatives: $12x^2y^2 + 8x^3y\frac{dy}{dx} - 20xy - 10x^2\frac{dy}{dx} = 0.$

3. Rearrange the equation to isolate $\frac{dy}{dx}$: $(8x^3y - 10x^2)\frac{dy}{dx} = -12x^2y^2 + 20xy.$

4. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-12x^2y^2 + 20xy}{8x^3y - 10x^2}.$

Thus, the final result is:

$\frac{dy}{dx} = \frac{20xy - 12x^2y^2}{8x^3y - 10x^2}.$

If you would like further details or have any questions, feel free to ask! Here are some related questions to consider:

1. What is the geometric interpretation of implicit differentiation?
2. Can you provide a specific example of where implicit differentiation is useful?
3. How would you solve for $y$ explicitly from the given equation?
4. What happens to the derivative if you change the equation slightly?
5. Are there other methods to find derivatives of implicitly defined functions?

Tip: When applying implicit differentiation, always remember to differentiate each term carefully, paying attention to the product rule for terms involving both $x$ and $y$.