To perform implicit differentiation on the equation:

$x^2y^3 + 4xy + x - 6y = 2$

we need to differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ (i.e., $y = y(x)$). We'll apply the product rule and chain rule where necessary. Let's differentiate each term:

Step 1: Differentiate the first term $x^2y^3$

We use the product rule, which states: $\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$ So, for $x^2y^3$: $\frac{d}{dx}(x^2y^3) = \frac{d}{dx}(x^2) \cdot y^3 + x^2 \cdot \frac{d}{dx}(y^3)$ Differentiating gives: $2x \cdot y^3 + x^2 \cdot 3y^2 \frac{dy}{dx}$ This simplifies to: $2x y^3 + 3x^2 y^2 \frac{dy}{dx}$

Step 2: Differentiate the second term $4xy$

Again using the product rule: $\frac{d}{dx}(4xy) = \frac{d}{dx}(4x) \cdot y + 4x \cdot \frac{d}{dx}(y)$ Differentiating gives: $4y + 4x \frac{dy}{dx}$

Step 3: Differentiate the third term $x$

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

Step 4: Differentiate the fourth term $-6y$

Since $y$ is a function of $x$, we apply the chain rule: $\frac{d}{dx}(-6y) = -6 \frac{dy}{dx}$

Step 5: Differentiate the constant on the right-hand side

$\frac{d}{dx}(2) = 0$

Step 6: Combine all differentiated terms

Now, combining everything:

$(2x y^3 + 3x^2 y^2 \frac{dy}{dx}) + (4y + 4x \frac{dy}{dx}) + 1 - 6 \frac{dy}{dx} = 0$

Step 7: Solve for $\frac{dy}{dx}$

Group the terms involving $\frac{dy}{dx}$:

$3x^2 y^2 \frac{dy}{dx} + 4x \frac{dy}{dx} - 6 \frac{dy}{dx} = - (2x y^3 + 4y + 1)$

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

$\frac{dy}{dx}(3x^2 y^2 + 4x - 6) = -(2x y^3 + 4y + 1)$

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

$\frac{dy}{dx} = \frac{-(2x y^3 + 4y + 1)}{3x^2 y^2 + 4x - 6}$

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

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Would you like more details or have any questions on this?

Here are 5 related questions for further practice:

1. How do you apply implicit differentiation when both variables are involved in trigonometric functions?
2. Can you explain how to use implicit differentiation on the equation $\sin(xy) + x^2 = y$?
3. What happens if you differentiate $x^2 + y^2 = 25$ implicitly?
4. How would you solve an implicit differentiation problem if it involves higher powers of $y$?
5. How do you interpret the result of implicit differentiation geometrically?

Tip: When using implicit differentiation, always treat $y$ as a function of $x$, and apply the product rule and chain rule where needed!